Matrix (二分套二分

Given a N × N matrix A, whose element in the i-th row and j-th column A_ij is an number that equals i² + 100000 × i + j² - 100000 × j + i × j, you are to find the M-th smallest element in the matrix.

Input

The first line of input is the number of test case.
For each test case there is only one line contains two integers, N(1 ≤ N ≤ 50,000) and M(1 ≤ M ≤ N × N). There is a blank line before each test case.

Output

For each test case output the answer on a single line.

Sample Input

12

1 1

2 1

2 2

2 3

2 4

3 1

3 2

3 8

3 9

5 1

5 25

5 10

Sample Output

3
-99993
3
12
100007
-199987
-99993
100019
200013
-399969
400031
-99939

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int t;
long long n,k;
long long make(long long i,long long j){
    return i*i+j*j+i*-j*+i*j;
}
bool erfen(long long m){
    long long cnt=;
    for(int j=;j<=n;j++){
        int l=,r=n,ans=;
        while(l<=r){//内层二分
            int i=l+r>>;
            if(make(i,j)<=m){
                ans=i;
                l=i+;
            }
            else r=i-;
        }
        cnt+=ans;
    }
    return cnt>=k;
}
int main(){
    cin>>t;
    for(int w=;w<t;w++){
        cin>>n>>k;
        long long l=-*n;
        long long r=n*n+n*n+*n+n*n,ans;
        while(l<=r){//外层二分
            long long m=l+r>>;
            if(erfen(m)){
                ans=m;
                r=m-;
            }else l=m+;
        }
        cout<<ans<<endl;
    }
    return ;
}