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 ab[100000000];
int ab1[1000000];
map<unsigned long long int, unsigned long long int> m;
map<unsigned long long int, unsigned long long int> m1;
map<unsigned long long int, unsigned long long int> m2;
vector<unsigned long long int> v;
vector<unsigned long long int> v1;
vector<unsigned long long int>::iterator itr;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string s12;
  long long int a, b, d = 100000000000000, j = 0, i = 0, t, s = 0, s2, as, as1,
                      n = 50, c = 0, c2 = 1, l;
  cin >> t;
  while (t--) {
    cin >> a >> b;
    if (a > b) {
      cout << b << endl;
    } else if (a == b) {
      cout << b + 1 << endl;
    } else {
      if (b % (a - 1) == 0) {
        c = 1;
      }
      cout << b + (b / (a - 1)) - c << endl;
      c = 0;
    }
  }
}