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;
inline long long toint(const std::string &s) {
  std::stringstream ss;
  ss << s;
  long long x;
  ss >> x;
  return x;
}
inline std::string tostring(long long number) {
  std::stringstream ss;
  ss << number;
  return ss.str();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  for (long long int tt = 0; tt < t; tt++) {
    long long int n, k;
    cin >> n >> k;
    if (k % (n - 1) == 0) {
      long long int x = k / (n - 1);
      x *= n;
      cout << x - 1 << "\n";
    } else {
      long long int x = k / (n - 1);
      x *= n;
      long long int y = k % (n - 1);
      cout << x + y << "\n";
    }
  }
  return 0;
}