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;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pil = pair<int, ll>;
using pll = pair<ll, ll>;
template <typename T>
using vec = vector<T>;
using vi = vec<int>;
using vl = vec<ll>;
template <typename T>
using que = queue<T>;
template <typename T>
using deq = deque<T>;
template <typename T>
T id(T b) {
  return b;
};
template <typename T>
void chmax(T &x, T y) {
  if (x < y) x = y;
}
template <typename T>
void chmin(T &x, T y) {
  if (x > y) x = y;
}
template <typename S, typename K>
bool contains(S &s, K k) {
  return s.find(k) != s.end();
}
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
constexpr ll TEN(int n) {
  if (n == 0)
    return 1LL;
  else
    return 10LL * TEN(n - 1);
}
int main() {
  fastio();
  int t;
  cin >> t;
  while (t--) {
    ll n, k;
    cin >> n >> k;
    ll l = 0, r = 3LL * TEN(9);
    while (r - l > 1) {
      ll m = (l + r) / 2;
      ll c = m / n;
      if (m - c < k) {
        l = m;
      } else {
        r = m;
      }
    }
    cout << r << "\n";
  }
  return 0;
}