#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    if (k < n) {
      cout << k << "\n";
    } else {
      long long rem = k / n;
      long long a;
      k % n == 0 ? a = k + n : a = ceil(float(k) / float(n)) * n;
      while (rem >= 0) {
        if (rem < n) {
          if ((a - k - 1) < rem) {
            cout << (a + (rem - (a - k - 1))) << "\n";
            break;
          } else {
            cout << (k + rem) << "\n";
            break;
          }
        } else {
          k += n * (rem / n);
          a += n * (rem / n);
          rem = rem % n + rem / n;
        }
      }
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    if (k < n) {
      cout << k << "\n";
    } else {
      long long rem = k / n;
      long long a;
      k % n == 0 ? a = k + n : a = ceil(float(k) / float(n)) * n;
      while (rem >= 0) {
        if (rem < n) {
          if ((a - k - 1) < rem) {
            cout << (a + (rem - (a - k - 1))) << "\n";
            break;
          } else {
            cout << (k + rem) << "\n";
            break;
          }
        } else {
          k += n * (rem / n);
          a += n * (rem / n);
          rem = rem % n + rem / n;
        }
      }
    }
  }
  return 0;
}

```