#include <bits/stdc++.h>
using namespace std;
bool SortbySecDesc(const pair<long long, long long> &a,
                   const pair<long long, long long> &b) {
  return a.second > b.second;
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    long long low = 1, high = 1e18, ans;
    while (low <= high) {
      long long mid = (low + high) / 2;
      long long vag = mid / a;
      long long baki = mid - vag;
      if (baki >= b) {
        ans = mid;
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
bool SortbySecDesc(const pair<long long, long long> &a,
                   const pair<long long, long long> &b) {
  return a.second > b.second;
}
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.first > b.first);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long a, b;
    cin >> a >> b;
    long long low = 1, high = 1e18, ans;
    while (low <= high) {
      long long mid = (low + high) / 2;
      long long vag = mid / a;
      long long baki = mid - vag;
      if (baki >= b) {
        ans = mid;
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```