#include <bits/stdc++.h>
using namespace std;
int t;
long long n;
int k;
int a;
long long ans;
long long pow1(long long x, int y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x) % 1000000007;
    ;
    x *= x;
    x %= 1000000007;
    ;
    y >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%d", &n, &k);
    a = 0;
    ans = 0;
    while (k) {
      for (int i = 32; i; i--) {
        if ((1 << i) > k && k >= (1 << (i - 1))) {
          a = i - 1;
          break;
        }
      }
      k -= (1 << a);
      ans += pow1(n, a);
      ans %= 1000000007;
      ;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### 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 t;
long long n;
int k;
int a;
long long ans;
long long pow1(long long x, int y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res * x) % 1000000007;
    ;
    x *= x;
    x %= 1000000007;
    ;
    y >>= 1;
  }
  return res;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%lld%d", &n, &k);
    a = 0;
    ans = 0;
    while (k) {
      for (int i = 32; i; i--) {
        if ((1 << i) > k && k >= (1 << (i - 1))) {
          a = i - 1;
          break;
        }
      }
      k -= (1 << a);
      ans += pow1(n, a);
      ans %= 1000000007;
      ;
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```