Number of pairs from two array with kth bit set on XOR

https://practice.geeksforgeeks.org/contest/job-a-thon-18-hiring-challenge/problems/#

Given two arrays A1, A2 both of size N, Q queries. Each query contains 5 elements, [k,l1,l2,r1,r2], where l1, l2 is the 1 based indexed range of A1, and r1, r2 is the 1 based indexed range of A2. Task is to find number of pairs such that A1[i]^A2[j] has the kth bit set, where l1<=i<=l2 and r1<=j<=r2.
Input
A1={1,2,3,4,5}
A2={1,2,3,4,5}
Queries=[[2,2,4,1,3] , [1,1,1,3,5]]
Output
4 1
Explaination
For 1st query
Segment from A1->{2,3,4}
Segment from A2->{1,2,3}
Possible pairs (arr[i],arr[j])=(2,1), (3,1), (4,2), (4,3) are the only pairs who have kth bit set when pair is XORed.
For 2nd query
Only pair will be (1,4)


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class Solution {
  public:
    vector<long long int> xorPairs(int N, vector<int> &A1, vector<int> &A2, int Q, vector<vector<int>> &query) {
        vector<vector<pair<long long,long long>>> dp1(N,vector<pair<long long,long long>>(31,{0,0}));  // zero,ones bit set
        vector<vector<pair<long long,long long>>> dp2(N,vector<pair<long long,long long>>(31,{0,0}));

        for(int i=0;i<N;i++){
            for(int bit=0;bit<=30;bit++){
                if(A1[i] & (1<<bit)){
                    dp1[i][bit].second++;
                }
                else{
                    dp1[i][bit].first++;
                }

                if(A2[i] & (1<<bit)){
                    dp2[i][bit].second++;
                }
                else{
                    dp2[i][bit].first++;
                }

                if(i-1>=0){
                    dp1[i][bit].second+=dp1[i-1][bit].second;
                    dp1[i][bit].first+=dp1[i-1][bit].first;

                    dp2[i][bit].second+=dp2[i-1][bit].second;
                    dp2[i][bit].first+=dp2[i-1][bit].first;
                }
            }
        }

        vector<long long int> res;

        for(auto q: query){
            int k=q[0]-1;
            int l1=q[1]-1;
            int l2=q[2]-1;
            int r1=q[3]-1;
            int r2=q[4]-1;

            long long zerosL=dp1[l2][k].first;
            long long onesL=dp1[l2][k].second;
            if(l1-1>=0){
                zerosL-=dp1[l1-1][k].first;
                onesL-=dp1[l1-1][k].second;
            }

            long long zerosR=dp2[r2][k].first;
            long long onesR=dp2[r2][k].second;
            if(r1-1>=0){
                onesR-=dp2[r1-1][k].second;
                zerosR-=dp2[r1-1][k].first;
            }

            long long int ans=onesR*zerosL+zerosR*onesL;
            res.push_back(ans);
        }
        return res;
    }
};