#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long fast_pow(long long n, long long k) {
  if (k == 0) {
    return 1;
  } else if (k % 2 == 1) {
    long long res = fast_pow(n, k / 2) % mod;
    return ((res * res) % mod * n) % mod;
  } else {
    long long res = fast_pow(n, k / 2) % mod;
    return (res * res) % mod;
  }
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    bitset<32> num(k);
    long long res = 0;
    for (long long i = 31; i >= 0; --i) {
      if (num[i] == 1) {
        res += fast_pow(n, i);
        res %= mod;
      }
    }
    cout << res % mod << endl;
  }
}

### 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;
long long mod = 1e9 + 7;
long long fast_pow(long long n, long long k) {
  if (k == 0) {
    return 1;
  } else if (k % 2 == 1) {
    long long res = fast_pow(n, k / 2) % mod;
    return ((res * res) % mod * n) % mod;
  } else {
    long long res = fast_pow(n, k / 2) % mod;
    return (res * res) % mod;
  }
}
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    bitset<32> num(k);
    long long res = 0;
    for (long long i = 31; i >= 0; --i) {
      if (num[i] == 1) {
        res += fast_pow(n, i);
        res %= mod;
      }
    }
    cout << res % mod << endl;
  }
}

```