#include <bits/stdc++.h>
using namespace std;
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int main() {
  fastio();
  long long int T = 1;
  cin >> T;
  for (long long int test = 1; test <= T; test++) {
    long long int n, m, k, ans, cnt, sum;
    ans = 0;
    cnt = 100;
    sum = 0;
    cin >> n >> k;
    if (k < n)
      ans = k;
    else {
      while (k && cnt--) {
        if (k < n) {
          ans += k;
          break;
        }
        long long int z = k / n;
        long long int a = z * n;
        z *= (n - 1);
        long long int q = k - z;
        ans += a;
        k = q;
      }
    }
    cout << ans << endl;
  }
}

### 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;
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
}
int main() {
  fastio();
  long long int T = 1;
  cin >> T;
  for (long long int test = 1; test <= T; test++) {
    long long int n, m, k, ans, cnt, sum;
    ans = 0;
    cnt = 100;
    sum = 0;
    cin >> n >> k;
    if (k < n)
      ans = k;
    else {
      while (k && cnt--) {
        if (k < n) {
          ans += k;
          break;
        }
        long long int z = k / n;
        long long int a = z * n;
        z *= (n - 1);
        long long int q = k - z;
        ans += a;
        k = q;
      }
    }
    cout << ans << endl;
  }
}

```