#include <bits/stdc++.h>
using namespace std;
long long large = 1000000007;
long long i, j;
long long dp[32000];
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void qwerty(void) {
  long long n, k;
  cin >> n >> k;
  long long t = k / (n - 1);
  long long x = k % (n - 1);
  long long ans = n * t + x;
  if (x == 0) {
    ans--;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test_cases;
  test_cases = 1;
  cin >> test_cases;
  for (i = 0; i < 32000; i++) {
    dp[i] = (((i) * (i + 1)) / 2) + ((i + 1) * (i + 2));
  }
  while (test_cases--) {
    qwerty();
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
long long large = 1000000007;
long long i, j;
long long dp[32000];
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void qwerty(void) {
  long long n, k;
  cin >> n >> k;
  long long t = k / (n - 1);
  long long x = k % (n - 1);
  long long ans = n * t + x;
  if (x == 0) {
    ans--;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test_cases;
  test_cases = 1;
  cin >> test_cases;
  for (i = 0; i < 32000; i++) {
    dp[i] = (((i) * (i + 1)) / 2) + ((i + 1) * (i + 2));
  }
  while (test_cases--) {
    qwerty();
  }
  return 0;
}

```