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>
int kq(int n, int k) {
  int kq;
  if (k % (n - 1) != 0)
    kq = n * (k / (n - 1)) + k % (n - 1);
  else
    kq = n * (k / (n - 1)) - 1;
  return kq;
}
int main() {
  int i, t, a[1000][2];
  scanf("%d", &t);
  for (i = 0; i < t; i++) {
    scanf("%d %d", &a[i][0], &a[i][1]);
  }
  for (i = 0; i < t; i++) {
    printf("%d\n", kq(a[i][0], a[i][1]));
  }
  return 0;
}