#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const long long LINF = 1e18;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long temp = n;
  long long po = 1;
  long long ans = 0;
  for (int bit = 0; bit < 31; bit++) {
    if ((k >> bit) & 1) {
      ans = ans + po;
      ans %= MOD;
    }
    po *= n;
    po %= MOD;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  cout << setprecision(10);
  int tc = 1;
  cin >> tc;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

Theofanis really likes sequences of positive integers, thus his teacher (Yeltsa Kcir) gave him a problem about a sequence that consists of only special numbers.

Let's call a positive number special if it can be written as a sum of different non-negative powers of n. For example, for n = 4 number 17 is special, because it can be written as 4^0 + 4^2 = 1 + 16 = 17, but 9 is not.

Theofanis asks you to help him find the k-th special number if they are sorted in increasing order. Since this number may be too large, output it modulo 10^9+7.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

The first and only line of each test case contains two integers n and k (2 ≤ n ≤ 10^9; 1 ≤ k ≤ 10^9).

Output

For each test case, print one integer — the k-th special number in increasing order modulo 10^9+7.

Example

Input


3
3 4
2 12
105 564


Output


9
12
3595374

Note

For n = 3 the sequence is [1,3,4,9...]

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 1;
const int MOD = 1e9 + 7;
const int INF = 1e9;
const long long LINF = 1e18;
void solve() {
  long long n, k;
  cin >> n >> k;
  long long temp = n;
  long long po = 1;
  long long ans = 0;
  for (int bit = 0; bit < 31; bit++) {
    if ((k >> bit) & 1) {
      ans = ans + po;
      ans %= MOD;
    }
    po *= n;
    po %= MOD;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed;
  cout << setprecision(10);
  int tc = 1;
  cin >> tc;
  for (int t = 1; t <= tc; t++) {
    solve();
  }
  return 0;
}

```