#include <bits/stdc++.h>
using namespace std;
long long t, n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> t;
  for (int i = 0; i < t; ++i) {
    cin >> n >> k;
    long long l = 1, r = 2000000001;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (mid - k == mid / n) {
        while (mid - k == mid / n) mid--;
        cout << mid + 1 << '\n';
        break;
      } else if (mid - k > mid / n)
        r = mid - 1;
      else
        l = mid + 1;
    }
  }
}

### Prompt

Please formulate a Cpp solution to 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;
long long t, n, k;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> t;
  for (int i = 0; i < t; ++i) {
    cin >> n >> k;
    long long l = 1, r = 2000000001;
    while (l <= r) {
      long long mid = (l + r) / 2;
      if (mid - k == mid / n) {
        while (mid - k == mid / n) mid--;
        cout << mid + 1 << '\n';
        break;
      } else if (mid - k > mid / n)
        r = mid - 1;
      else
        l = mid + 1;
    }
  }
}

```