#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t, n, k;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    scanf("%lld", &k);
    long long low = 1;
    long long high = n * k;
    long long ans = -1;
    while (low <= high) {
      long long mid = low + (high - low) / 2;
      long long x = mid - (mid / n);
      if (x > k)
        high = mid - 1;
      else if (x < k)
        low = mid + 1;
      else if (x == k) {
        if (mid % n != 0) {
          ans = mid;
          break;
        } else
          high = mid - 1;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long t, n, k;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    scanf("%lld", &k);
    long long low = 1;
    long long high = n * k;
    long long ans = -1;
    while (low <= high) {
      long long mid = low + (high - low) / 2;
      long long x = mid - (mid / n);
      if (x > k)
        high = mid - 1;
      else if (x < k)
        low = mid + 1;
      else if (x == k) {
        if (mid % n != 0) {
          ans = mid;
          break;
        } else
          high = mid - 1;
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```