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 int Int() {
  int x;
  cin >> x;
  return x;
}
inline long long Long() {
  long long x;
  cin >> x;
  return x;
}
inline float Float() {
  float x;
  cin >> x;
  return x;
}
inline double Double() {
  double x;
  cin >> x;
  return x;
}
inline void Yes() { cout << "Yes" << endl; }
inline void No() { cout << "No" << endl; }
inline void YES() { cout << "YES" << endl; }
inline void NO() { cout << "NO" << endl; }
const int N = (int)2e5 + 5;
const int maxN = (int)1e6 + 6;
const long long Mod = (long long)1e9 + 7;
const int inf = (int)2e9;
const long long Inf = (long long)1e18;
int main() {
  int t, q;
  cin >> t;
  for (q = 1; q <= t; q++) {
    long long N, K;
    cin >> N >> K;
    if (K % (N - 1) == 0)
      cout << (K / (N - 1)) * N - 1 << endl;
    else
      cout << (K / (N - 1)) * N + K % (N - 1) << endl;
  }
  return 0;
}