#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int MOD = 1e9 + 7;
    long long int ans = 0;
    long long int ctr = 0;
    while (k != 0) {
      if (k % 2 == 1) {
        ans = (ans + power(n, ctr, MOD) % MOD) % MOD;
      } else {
      }
      ctr++;
      k /= 2;
    }
    cout << ans % MOD << endl;
  }
}

### 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;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, k;
    cin >> n >> k;
    long long int MOD = 1e9 + 7;
    long long int ans = 0;
    long long int ctr = 0;
    while (k != 0) {
      if (k % 2 == 1) {
        ans = (ans + power(n, ctr, MOD) % MOD) % MOD;
      } else {
      }
      ctr++;
      k /= 2;
    }
    cout << ans % MOD << endl;
  }
}

```