Time complexity: $$$O(N^2+QlogN)$$$

I made a knapsack $$$ks$$$ as an array of size of $$$2 \cdot N$$$ , which $$$ks_i$$$ represent the minimum cost to get $$$i$$$ amount of points.(takes time of $$$O(N^2)$$$ of precomputing), then I received the queries and binary searched the maximum amount of points we can get and outputted it.(takes time of $$$O(QlogN)$$$) aaaand I'm stuck here. I don't know how to optimize this further...

/********************
 * what  the  sigma *
 ********************/
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <algorithm>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
using namespace std;
#define lgm cin.tie(0)->sync_with_stdio(0);
#define be(x) x.begin(),x.end()
#define ve vector
#define ll long long
#define ld long double
#define db(x) cout << "Debug at " << x << endl;
#define ull unsigned ll
#define f first
#define s second
#define pii pair<int, int>
#define tii tuple<int,int,int>
#define pll pair<ll,ll>
#define sz(x) (int)x.size()
#define pb push_back
const int mod = 1e9+7,maxn=200005;
ll ks[200001];
int main() {
    cin.tie(0)->sync_with_stdio(0);
    ll n,cnt=0,cnt2=0;
    cin >> n;
    ll a[n],b[n];
    vector<ll> v;
    for (ll i=0;i<n;i++) {
        cin >> a[i];
        v.push_back(a[i]);
    }
    for (ll i=0;i<n;i++) {
        cin >> b[i];
        v.push_back(b[i]);
        if (a[i] <= b[i]) {
            cnt++;
        } else if (a[i]>=b[i]) {
            cnt2++;
        }
    }
    ll q;
    cin >> q;
    ll m=100000;
    m=m*m*m;
    ll x;
    for (int i=1;i<=2*n;i++) {
        ks[i]=m;
    }
    for (int i=0;i<n;i++) {
        for (int j=2*(i+1);j>=2;j--) {
            ks[j]=min(ks[j],min(ks[j-1]+a[i],ks[j-2]+a[i]+b[i]));
        }
        ks[1]=min(a[i],ks[1]);
    }
    while (q--) {
        cin >> x;
        int l=upper_bound(ks,ks+2*n+1,x)-ks-1;
        cout << l<<'\n';
    }
}