#include <bits/stdc++.h>
using namespace std;
int n, k, t;
map<int, int> m;
long long s, ss = 1;
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 0; i >= 0; i++) {
      if (k == 1) {
        m[i] = 1;
        for (int j = 1; j <= i; j++) {
          ss *= n;
          ss %= 1000000007;
        }
        s += ss;
        ss = 1;
        break;
      }
      if (k % 2 == 1) {
        m[i] = 1;
        for (int j = 1; j <= i; j++) {
          ss *= n;
          ss %= 1000000007;
        }
        s += ss;
        ss = 1;
      }
      k /= 2;
    }
    s %= 1000000007;
    cout << s << '\n';
    s = 0;
  }
}

### Prompt

Create a solution in CPP for the following problem:

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;
int n, k, t;
map<int, int> m;
long long s, ss = 1;
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> k;
    for (int i = 0; i >= 0; i++) {
      if (k == 1) {
        m[i] = 1;
        for (int j = 1; j <= i; j++) {
          ss *= n;
          ss %= 1000000007;
        }
        s += ss;
        ss = 1;
        break;
      }
      if (k % 2 == 1) {
        m[i] = 1;
        for (int j = 1; j <= i; j++) {
          ss *= n;
          ss %= 1000000007;
        }
        s += ss;
        ss = 1;
      }
      k /= 2;
    }
    s %= 1000000007;
    cout << s << '\n';
    s = 0;
  }
}

```