#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2) res = (res * a) % 1000000007;
    b /= 2;
    a = (a * a) % 1000000007;
  }
  return res;
}
void solve() {
  long long n, m, i, j, k, l, sum = 1, flag = 0, t, ans = 0;
  string s, s1 = "", s2;
  cin >> n >> k;
  i = k / (n - 1);
  j = k % (n - 1);
  if (j == 0) j = -1;
  cout << (n * i + j) << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  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;
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2) res = (res * a) % 1000000007;
    b /= 2;
    a = (a * a) % 1000000007;
  }
  return res;
}
void solve() {
  long long n, m, i, j, k, l, sum = 1, flag = 0, t, ans = 0;
  string s, s1 = "", s2;
  cin >> n >> k;
  i = k / (n - 1);
  j = k % (n - 1);
  if (j == 0) j = -1;
  cout << (n * i + j) << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int test = 1;
  cin >> test;
  while (test--) {
    solve();
  }
  return 0;
}

```