#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
void solve() {
  ll n, k;
  cin >> n >> k;
  ll x = n * (k / (n - 1)) - 1;
  k -= (n - 1) * (k / (n - 1));
  if (k == 0) {
    cout << x << endl;
    return;
  }
  x += 1 + k;
  cout << x << endl;
  return;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int q;
  cin >> q;
  while (q--) solve();
  return 0;
}

### Prompt

Generate a cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll LINF = 1001002003004005006ll;
int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
void solve() {
  ll n, k;
  cin >> n >> k;
  ll x = n * (k / (n - 1)) - 1;
  k -= (n - 1) * (k / (n - 1));
  if (k == 0) {
    cout << x << endl;
    return;
  }
  x += 1 + k;
  cout << x << endl;
  return;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int q;
  cin >> q;
  while (q--) solve();
  return 0;
}

```