Implicit 2d fenwick tree should be enough I guess These requests are just +1 in point and sum on prefix

unordered_map<int, unordered_map<int, int>> t; // or smth like that

int get(int x, int y) {
    int res = 0;
    for (int i = x; i > 0; i -= i & -i)
        for (int j = y; j > 0; j -= j & -j)
            res += t[i][j];
    return res;
}

void add(int x, int y, int v) {
    for (int i = x; i <= MAX; i += i & -i)
        for (int j = y; j <= MAX; j += j & -j)
            t[i][j] += v;
}

will be almost the whole code. Not sure where you can read about it, the first blog i found: https://mirror.codeforces.com/blog/entry/57292

My bad lol

WRONG

An obvious idea is to make 2 multisets of points, one sorted by $$$x$$$ and another by $$$y$$$. For each 2nd type query we find the intersection of multisets, one of which has points which satisfie $$$x_i \le x$$$, and the second $$$y_i \le y$$$, both of which can be computed in $$$O(log(n))$$$. However, this depends on the intersection finding time complexity, which is unknown for me.

Time complexity for each query is $$$O(log(n)*F(s,d))$$$, where $$$F(s,d)$$$ is time needed to compute the intersection of multisets $$$s$$$ and $$$d$$$

Hope this helps

This problem (statement only in Romanian) is very simmilar. You are asked for multiple values to compute how many values to the left and above are equal to that value and sum these counts. The solution is in $$$O(N^2 \cdot \log N)$$$ using the same idea sugested by others.

I didn't think much, but you should be able to modify the first example problem from here to solve your problem offline.