#include <bits/stdc++.h>
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int MOD = 1e9 + 7;
const long long SEED = 1333331;
const int N = 1e6 + 10, M = N + N;
using namespace std;
void solve() {
  long long n, m;
  cin >> n >> m;
  cout << m / (n - 1) * n + (m % (n - 1) == 0 ? -1 : m % (n - 1)) << endl;
  return;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
  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>
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int MOD = 1e9 + 7;
const long long SEED = 1333331;
const int N = 1e6 + 10, M = N + N;
using namespace std;
void solve() {
  long long n, m;
  cin >> n >> m;
  cout << m / (n - 1) * n + (m % (n - 1) == 0 ? -1 : m % (n - 1)) << endl;
  return;
}
int main() {
  int T;
  cin >> T;
  while (T--) solve();
  return 0;
}

```