#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % 1000000007;
      ;
      b--;
    } else {
      a = (a * a) % 1000000007;
      ;
      b /= 2;
    }
  }
  return res;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long ans = 0;
    vector<long long> v;
    while (k) {
      v.push_back(k % 2);
      k /= 2;
    }
    for (int i = 0; i < v.size(); i++) {
      if (v[i]) {
        ans = (ans + power(n, i)) % 1000000007;
        ;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % 1000000007;
      ;
      b--;
    } else {
      a = (a * a) % 1000000007;
      ;
      b /= 2;
    }
  }
  return res;
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long ans = 0;
    vector<long long> v;
    while (k) {
      v.push_back(k % 2);
      k /= 2;
    }
    for (int i = 0; i < v.size(); i++) {
      if (v[i]) {
        ans = (ans + power(n, i)) % 1000000007;
        ;
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```