#include <bits/stdc++.h>
using namespace std;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int low = 0;
    int high = INT_MAX;
    int mid = 0;
    while (low < high) {
      mid = low + (high - low) / 2;
      if ((mid - mid / n) > k)
        high = mid;
      else if ((mid - mid / n) < k)
        low = mid + 1;
      else
        break;
    }
    if (mid % n == 0) mid--;
    cout << mid << endl;
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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() {
  int t;
  cin >> t;
  while (t--) {
    int n, k;
    cin >> n >> k;
    int low = 0;
    int high = INT_MAX;
    int mid = 0;
    while (low < high) {
      mid = low + (high - low) / 2;
      if ((mid - mid / n) > k)
        high = mid;
      else if ((mid - mid / n) < k)
        low = mid + 1;
      else
        break;
    }
    if (mid % n == 0) mid--;
    cout << mid << endl;
  }
  return 0;
}

```