#include <bits/stdc++.h>
using namespace std;
using ll = long long;
constexpr int N = 1e6 + 5;
constexpr int MOD = 1e9 + 7;
int T;
int main() {
  scanf("%d", &T);
  while (T--) {
    int n, k;
    scanf("%d%d", &n, &k);
    ll d = 1;
    ll ret = 0;
    while (k) {
      if (k & 1) {
        ret = (ret + d) % MOD;
      }
      d = d * n % MOD;
      k >>= 1;
    }
    printf("%lld\n", ret);
  }
}

### 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;
using ll = long long;
constexpr int N = 1e6 + 5;
constexpr int MOD = 1e9 + 7;
int T;
int main() {
  scanf("%d", &T);
  while (T--) {
    int n, k;
    scanf("%d%d", &n, &k);
    ll d = 1;
    ll ret = 0;
    while (k) {
      if (k & 1) {
        ret = (ret + d) % MOD;
      }
      d = d * n % MOD;
      k >>= 1;
    }
    printf("%lld\n", ret);
  }
}

```