First of all, we can observe that queries can be done offline and the only thing that matters for each query is the color.

Lets do each color one by one, we can observe that distance between two nodes in the query = original distance $$$-$$$ sum of edge weight of color $$$x$$$ + number of edges that have color $$$x$$$ in the path * new weight length.

Now let us try to find what can be done. We can see that the number of edges with color $$$x$$$ going out of subtree $$$u$$$ will $$$+1$$$ if the starting point is from subtree $$$u$$$. So we can just $$$+1$$$ to all nodes inside subtree $$$u$$$. Then we can see that number of edges from node $$$x$$$ to $$$y$$$. Similarly, we can do $$$+W$$$ for weight sum.

Distance in original graph can be found in $$$O(N log N)$$$ using binary lifting and a dfs function. Simultaneously, since we need to add $$$+1$$$ to each node one by one, we will need to do $$$O(N ^ 2)$$$ to do all necessary operations.

We can try finding a better way to do the $$$+1$$$ and $$$+W$$$ operations. Well, if you have learnt dfn, you should be able to solve the problem from here. If we represent $$$tin_{ i }$$$ be the time we entered subtree $$$i$$$ and $$$tout_{ i }$$$ we left subtree $$$i$$$. Then we can use this to observe that we need to do $$$+val$$$ in a range and find a value at a point. What do we need to have if we want to do these operations fast? Yes, segment tree!

However, we cannot just declare a new segment tree each time. We need to reuse the segment tree. So, we'll have to do $$$-1$$$ and $$$-W$$$ operations to revert the segment tree, or do something like range set.

// LUOGU_RID: 139270322
// Problem: F - Colorful Tree
// Memory Limit: 1024 MB
// Time Limit: 4000 ms

//闲敲棋子落灯花//
#include<bits/stdc++.h>
using namespace std;
using i64 = long long;

struct edge {
    int u, v, color, w;
};
struct query {
    int u, v, color, new_w, id;
};
const int N = 1e5 + 50, K = 30;
vector<pair<int, int>> G[N];
int dep[N], sp[N][K], val[N], ti, tin[N], tout[N];
void dfs(int u, int p) {
    tin[u] = ++ti; dep[u] = dep[p] + 1; sp[u][0] = p;
    for (int i = 1; i < K; i++)sp[u][i] = sp[sp[u][i - 1]][i - 1];
    for (auto v : G[u]) if (v.first != p) {
        val[v.first] = val[u] + v.second;
        dfs(v.first, u);
    }
    tout[u] = ti;
}
int lca(int u, int v) {
    if (dep[u] < dep[v]) swap(u, v);
    for (int i = K - 1; i >= 0; i--)if (dep[sp[u][i]] >= dep[v])u = sp[u][i];
    if (u == v) return u;
    for (int i = K - 1; i >= 0; i--)if (sp[u][i] != sp[v][i])u = sp[u][i], v = sp[v][i];
    return sp[v][0];
}
int dis(int u, int v) {
    int l = lca(u, v);
    return val[u] + val[v] - 2 * val[l];
}
struct segtree {
    struct node {
        i64 v;
        friend node operator + (node a, node b) {
            node t; t.v = a.v + b.v;
            return t;
        }
    };
    vector<node> tr;
    vector<int> tag;
    int SZ = 0;
    void apply_node(int p, int x) {
        tr[p].v += x;
        tag[p] += x;
    }
    void pushdown(int p) {
        if (tag[p] != 0) {
            apply_node(p << 1, tag[p]);
            apply_node(p << 1 | 1, tag[p]);
            tag[p] = 0;
        }
        return;
    }

    segtree(int N) {
        SZ = N + 200;
        tr.resize(SZ * 4);
        tag.assign(SZ * 4, 0);
    }
#define ls p<<1
#define rs p<<1|1
    void update(int x, int v) { update(1, 1, SZ - 100, v, x, x); }
    void update(int x, int y, int v) { update(1, 1, SZ - 100, v, x, y); }
    void update(int p, int l, int r, int v, int lq, int rq) {
        //cout << "update: " << p << ' ' << l << ' ' << r << ' ' << lq << ' ' << rq << '\n';
        if (lq <= l && r <= rq) {
            apply_node(p, v);
            return;
        }
        int mid = (l + r) >> 1;
        pushdown(p);
        if (lq <= mid) update(ls, l, mid, v, lq, rq);
        if (rq >= mid + 1) update(rs, mid + 1, r, v, lq, rq);
        tr[p] = tr[ls] + tr[rs];
    }
    node query(int x) { return query(1, 1, SZ - 100, x, x); }
    node query(int x, int y) { return query(1, 1, SZ - 100, x, y); }
    node query(int p, int l, int r, int lq, int rq) {
        //cout << "query: " << p << ' ' << l << ' ' << r << ' ' << lq << ' ' << rq << '\n';
        if (lq <= l && r <= rq) {
            return tr[p];
        }
        int mid = (l + r) >> 1;
        pushdown(p);
        if (lq <= mid && rq >= mid + 1) return query(ls, l, mid, lq, rq) + query(rs, mid + 1, r, lq, rq);
        if (lq <= mid) return query(ls, l, mid, lq, rq);
        if (rq >= mid + 1) return query(rs, mid + 1, r, lq, rq);
    }
};
//update/query -> point or range
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    int N, Q; cin >> N >> Q;
    vector<vector<edge>> C(N + 1);
    for (int i = 0; i < N - 1; i++) {
        int a, b, c, d; cin >> a >> b >> c >> d;
        G[a].push_back({ b,d });
        G[b].push_back({ a,d });
        C[c].push_back({ a,b,c,d });
    }
    dfs(1, 0);
    vector<vector<query>> Qp(N + 1);
    vector<int> ans(Q + 1);
    for (int i = 0; i < Q; i++) {
        int a, b, c, d; cin >> a >> b >> c >> d;
        ans[i] = dis(c, d);
        Qp[a].push_back({ c,d,a,b,i });
    }
    /*
        let the initial answer for each query to be
        l = lca(u,v), ans[i] = dis(u,l) + dis(l,v)
        loop through each color:
            add the entire subtree  S       +1 when going down
                                    dep     +w when going down
            then ans[i] -= dis(a,b) on the subtree
                 ans[i] += S(a,b) * new_w

        this is O(N * C) where C = color count
        we cant just traverse every single node so lets use
        dfn values and a segment tree
    */
    segtree segcnt(N), segval(N);
    for (int i = 1; i <= N; i++) {
        for (auto p : C[i]) {
            int u = p.u, v = p.v, w = p.w;
            if (dep[u] > dep[v]) swap(u, v);
            segcnt.update(tin[v], tout[v], 1);
            segval.update(tin[v], tout[v], w);
        }

        for (auto p : Qp[i]) {
            int u = p.u, v = p.v, w = p.new_w;
            if (dep[u] > dep[v]) swap(u, v);
            int l = lca(u, v);
            int cnt = segcnt.query(tin[u]).v + segcnt.query(tin[v]).v - 2 * segcnt.query(tin[l]).v;
            int val = segval.query(tin[u]).v + segval.query(tin[v]).v - 2 * segval.query(tin[l]).v;
            ans[p.id] -= val;
            ans[p.id] += cnt * p.new_w;
        }

        for (auto p : C[i]) {
            int u = p.u, v = p.v, w = p.w;
            if (dep[u] > dep[v]) swap(u, v);
            segcnt.update(tin[v], tout[v], -1);
            segval.update(tin[v], tout[v], -w);
        }
    }
    for (int i = 0; i < Q; i++) {
        cout << ans[i] << '\n';
    }
}