#include <bits/stdc++.h>
using namespace std;
const double esp = 1e-6;
const long long N = 1e4 + 9;
const long long maxn = 2e5 + 9;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long d, y;
  d = k / (n - 1), y = k % (n - 1);
  if (y == 0) {
    cout << (d - 1) * n + n - 1 << '\n';
  } else {
    cout << (d)*n + y << '\n';
  }
}
signed main() {
  long long _;
  cin >> _;
  while (_--) solve();
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
const double esp = 1e-6;
const long long N = 1e4 + 9;
const long long maxn = 2e5 + 9;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long d, y;
  d = k / (n - 1), y = k % (n - 1);
  if (y == 0) {
    cout << (d - 1) * n + n - 1 << '\n';
  } else {
    cout << (d)*n + y << '\n';
  }
}
signed main() {
  long long _;
  cin >> _;
  while (_--) solve();
}

```