Given a N × N matrix A, whose element in the i-th row and j-th column Aij 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

题意：N阶矩阵Aij= i2 + 100000 × i + j2 – 100000 × j + i × j，求第M小的元素。

分析：首先你要知道这个式子是关于i递增的二元函数，如果是知道这一步，那剩下的就交给二分

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
using namespace std;
typedef long long ll;
#define rep(i,l,r) for(int i=l;i<=r;i++)
const int N = 3e5 + 10;
const ll INF = 0x3f3f3f3f3f3f3f;
ll m;
ll n;
ll Get(ll x,ll y){
    return x*x+100000*x+y*y-100000*y+y*x;
}
bool solve(ll x)//枚举小于当前x的个数有几个
{
    int tot=0;
    rep(j,1,n){//每一列去枚举
    int l=0,r=n+1;
    while(r-l>1)
    {
        int mid=(l+r)>>1;
        if(Get(mid,j)<x) l=mid;//找到最后在当前列下，有几行的元素是小于当前x的
        else r=mid;
    }
    tot+=l;//把满足的每列的行的个数相加
    }
    return tot<m;//如果小于当前x的数是小于m个，那就是说明x太小了，那就是需要增加下界
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif // ONLINE_JUDGE
    int t;
    scanf("%d",&t);
    while(t--){
    scanf("%lld%lld",&n,&m);
    ll lb = -INF, ub = INF;//枚举可能的值，因为数值很大，因此希望是从无穷小到无穷大之间选择
    while(ub-lb>1){
        ll mid=(lb+ub)>>1;
        if(solve(mid))
            lb=mid;
        else ub=mid;
    }
    printf("%lld\n",lb);
    }
    return 0;
}