#include <bits/stdc++.h>
using namespace std;
vector<int> a;
void decToBinary(long long n) {
  a.clear();
  while (n > 0) {
    a.push_back(n % 2);
    n = n / 2;
  }
}
int main() {
  const unsigned int M = 1000000007;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    decToBinary(k);
    unsigned long long sum = 0, po = 1;
    for (auto it = a.begin(); it != a.end(); it++) {
      if (*it == 0) {
        po = po * n;
        po = po % M;
      } else {
        sum = sum + po;
        po = po * n;
        po = po % M;
      }
    }
    sum = sum % M;
    cout << sum << "\n";
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
vector<int> a;
void decToBinary(long long n) {
  a.clear();
  while (n > 0) {
    a.push_back(n % 2);
    n = n / 2;
  }
}
int main() {
  const unsigned int M = 1000000007;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    decToBinary(k);
    unsigned long long sum = 0, po = 1;
    for (auto it = a.begin(); it != a.end(); it++) {
      if (*it == 0) {
        po = po * n;
        po = po % M;
      } else {
        sum = sum + po;
        po = po * n;
        po = po % M;
      }
    }
    sum = sum % M;
    cout << sum << "\n";
  }
}

```