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;
const int mod = (int)1000000000 + 7;
const int MOD = (int)998244353;
const int INF = (int)1e18;
bool cmp(const pair<int, int>& a, const pair<int, int>& b) {
  return a.second - a.first < b.second - b.first;
}
long long bit_ct1(long long x) {
  long long ct = 0;
  while (x > 0) {
    x = x & (x - 1);
    ct++;
  }
  return ct;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long p;
  p = (k) / (n - 1);
  if (k % (n - 1)) p++;
  k = (n * (p)-p) - k;
  cout << n * p - k - 1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  clock_t timest = clock();
  cerr << "[Executes in " << (double)(clock() - timest) / CLOCKS_PER_SEC << "s]"
       << '\n';
}