#include <bits/stdc++.h>
const int maxn = 1000000 + 50;
const int maxm = 4e5 + 10;
const int inf_max = 0x3f3f3f;
using namespace std;
long long n, k;
const long long MAX = 2000000000;
long long check(long long cur) {
  long long group = cur / n;
  cur -= group;
  return cur;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long l = 1, r = 3000000000;
    while (l <= r) {
      long long mid = (l + r) >> 1;
      if (mid % n == 0) {
        long long tt = mid / n, tmp = mid;
        tmp -= tt;
        if (tmp == k) {
          printf("%lld\n", mid - 1);
          break;
        } else if (tmp > k)
          r = mid - 1;
        else
          l = mid + 1;
      } else {
        long long cur = check(mid);
        if (cur == k) {
          printf("%lld\n", mid);
          break;
        } else if (cur > k)
          r = mid - 1;
        else
          l = mid + 1;
      }
    }
  }
  return 0;
}

### Prompt

Your task is to create 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>
const int maxn = 1000000 + 50;
const int maxm = 4e5 + 10;
const int inf_max = 0x3f3f3f;
using namespace std;
long long n, k;
const long long MAX = 2000000000;
long long check(long long cur) {
  long long group = cur / n;
  cur -= group;
  return cur;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    long long l = 1, r = 3000000000;
    while (l <= r) {
      long long mid = (l + r) >> 1;
      if (mid % n == 0) {
        long long tt = mid / n, tmp = mid;
        tmp -= tt;
        if (tmp == k) {
          printf("%lld\n", mid - 1);
          break;
        } else if (tmp > k)
          r = mid - 1;
        else
          l = mid + 1;
      } else {
        long long cur = check(mid);
        if (cur == k) {
          printf("%lld\n", mid);
          break;
        } else if (cur > k)
          r = mid - 1;
        else
          l = mid + 1;
      }
    }
  }
  return 0;
}

```