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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool possible(long long x, long long n, long long k) {
  if (x >= ceil((long double)(k + x) / ((long double)(n)))) {
    return true;
  }
  return false;
}
void solve() {
  long long n, k, x = 1;
  cin >> n >> k;
  long long left = 1, right = x, mid;
  while (left <= right) {
    mid = left + (right - left) / 2;
    if (possible(mid, n, k)) {
      right = mid - 1;
    } else {
      left = mid + 1;
    }
  }
  cout << " ["
       << "left"
          ": "
       << (left) << "] " << endl;
  ;
  x = left;
  cout << ((n * x) - 1) - ((n - 1) * x - k) << endl;
}
void solve_2() {
  long long n, k, x, ans;
  cin >> n >> k;
  x = (k) / (n - 1) + (k % (n - 1) == 0 ? 0 : 1);
  ans = n * x - (((n - 1) * x) - k) - 1;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tt;
  cin >> tt;
  while (tt--) {
    solve_2();
  }
}