#include <bits/stdc++.h>
using namespace std;
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long min(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
long long max(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main(int argc, char **argv) {
  long long t;
  cin >> t;
  for (long long ii = 0; ii < t; ii++) {
    long long n, k;
    cin >> n >> k;
    long long solution = 0;
    long long MOD = 1000000007;
    long long mult = 1;
    while (k > 0) {
      if (k % 2 == 1) {
        solution += mult;
      }
      mult *= n;
      mult %= MOD;
      k /= 2;
      solution %= MOD;
    }
    cout << solution << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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 max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long min(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
long long max(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
int main(int argc, char **argv) {
  long long t;
  cin >> t;
  for (long long ii = 0; ii < t; ii++) {
    long long n, k;
    cin >> n >> k;
    long long solution = 0;
    long long MOD = 1000000007;
    long long mult = 1;
    while (k > 0) {
      if (k % 2 == 1) {
        solution += mult;
      }
      mult *= n;
      mult %= MOD;
      k /= 2;
      solution %= MOD;
    }
    cout << solution << endl;
  }
  return 0;
}

```