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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1, r, n, a, d, c, b, x, y, sum, cnt, i, j, k, m, l, q, maxx,
                minn, temp, ans, flag, g, ts;
  char ch, chh;
  double dans;
  string str, ttr;
  cin >> t;
  for (ts = 1; ts <= t; ts++) {
    flag = 0;
    sum = 0;
    cnt = 0;
    ans = 0;
    k = 0;
    maxx = 0;
    minn = 0;
    a = 0;
    b = 0;
    temp = 0;
    cin >> n >> k;
    x = (k / n);
    temp = x;
    k += x;
    while (x != 0) {
      x = (k / n);
      x = (x - temp);
      temp = (k / n);
      k += x;
    }
    cout << k;
    cout << "\n";
  }
  return 0;
}