#include <bits/stdc++.h>
using namespace std;
int main() {
  const long long MOD = 1000000007;
  int t;
  cin >> t;
  for (int _t = 0; _t < t; _t++) {
    int n;
    cin >> n;
    int k;
    cin >> k;
    int binary_k[32];
    int powers_of_two[32];
    powers_of_two[0] = 1;
    for (int i = 1; i < 32; i++) {
      powers_of_two[i] = powers_of_two[i - 1] * 2;
    }
    for (int i = 31; i >= 0; i--) {
      binary_k[i] = k / powers_of_two[i];
      k %= powers_of_two[i];
    }
    long long counter = 0;
    long long kth_power_of_n = 1;
    for (int i = 0; i < 32; i++) {
      counter += binary_k[i] * kth_power_of_n;
      counter %= MOD;
      kth_power_of_n *= n;
      kth_power_of_n %= MOD;
    }
    cout << counter << endl;
  }
}

### 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;
int main() {
  const long long MOD = 1000000007;
  int t;
  cin >> t;
  for (int _t = 0; _t < t; _t++) {
    int n;
    cin >> n;
    int k;
    cin >> k;
    int binary_k[32];
    int powers_of_two[32];
    powers_of_two[0] = 1;
    for (int i = 1; i < 32; i++) {
      powers_of_two[i] = powers_of_two[i - 1] * 2;
    }
    for (int i = 31; i >= 0; i--) {
      binary_k[i] = k / powers_of_two[i];
      k %= powers_of_two[i];
    }
    long long counter = 0;
    long long kth_power_of_n = 1;
    for (int i = 0; i < 32; i++) {
      counter += binary_k[i] * kth_power_of_n;
      counter %= MOD;
      kth_power_of_n *= n;
      kth_power_of_n %= MOD;
    }
    cout << counter << endl;
  }
}

```