#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5, inf = 2e9, mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int tc, n, k;
  long long low, high, mid;
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    low = 1, high = 2e18;
    while (high >= low) {
      mid = (high + low) / 2;
      if (mid - mid / n > k)
        high = mid - 1;
      else if (mid - mid / n < k)
        low = mid + 1;
      else {
        cout << mid - (mid % n == 0) << '\n';
        break;
      }
    }
  }
  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;
const int N = 1e5 + 5, inf = 2e9, mod = 1e9 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  int tc, n, k;
  long long low, high, mid;
  cin >> tc;
  while (tc--) {
    cin >> n >> k;
    low = 1, high = 2e18;
    while (high >= low) {
      mid = (high + low) / 2;
      if (mid - mid / n > k)
        high = mid - 1;
      else if (mid - mid / n < k)
        low = mid + 1;
      else {
        cout << mid - (mid % n == 0) << '\n';
        break;
      }
    }
  }
  return 0;
}

```