Problem: You are given two positive integers n and k. Print the k-th positive integer that is not divisible by n.

For example, if n=3, and k=7, then all numbers that are not divisible by 3 are: 1, 2, 4, 5, 7, 8, 10, 11, 13 .... The 7-th number among them is 10.

Input

The first line contains an integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. Next, t test cases are given, one per line.

Each test case is two positive integers n (2 ≤ n ≤ 10^9) and k (1 ≤ k ≤ 10^9).

Output

For each test case print the k-th positive integer that is not divisible by n.

Example

Input


6
3 7
4 12
2 1000000000
7 97
1000000000 1000000000
2 1


Output


10
15
1999999999
113
1000000001
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
int close_sycn = (ios::sync_with_stdio(false), cin.tie(0), 0);
int main() {
  LL t, n, k;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    LL res = k / (n - 1) * n;
    LL p = k % (n - 1);
    res += p == 0 ? -1 : p;
    cout << res << '\n';
  }
  return 0;
}