#include <bits/stdc++.h>
using namespace std;
bool cmp(pair<long long, long long> p1, pair<long long, long long> p2) {
  if (p1.first == p2.first)
    return p1.second > p2.second;
  else
    return p1.first < p2.first;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long high = 3 * (long long)pow(10, 9), low = 1, ans = -1;
    while (low <= high) {
      long long mid = (low + high) / 2;
      if ((mid - ((long long)mid / n)) == k) {
        long long j = mid;
        while ((j - ((long long)j / n)) == k) {
          j--;
        }
        ans = j + 1;
        break;
      } else if ((mid - ((long long)mid / n)) > k) {
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    }
    cout << ans << 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;
bool cmp(pair<long long, long long> p1, pair<long long, long long> p2) {
  if (p1.first == p2.first)
    return p1.second > p2.second;
  else
    return p1.first < p2.first;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long high = 3 * (long long)pow(10, 9), low = 1, ans = -1;
    while (low <= high) {
      long long mid = (low + high) / 2;
      if ((mid - ((long long)mid / n)) == k) {
        long long j = mid;
        while ((j - ((long long)j / n)) == k) {
          j--;
        }
        ans = j + 1;
        break;
      } else if ((mid - ((long long)mid / n)) > k) {
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```