#include <bits/stdc++.h>
using namespace std;
long long m = 1e9 + 7;
long long n, k, aa[10001], sum, cnt;
void solve() {
  cin >> n >> k;
  aa[0] = 1;
  for (int i = 1; i <= 32; i++) {
    aa[i] = ((aa[i - 1] % m) * (n % m)) % m;
  }
  while (k) {
    if (k % 2 != 0) {
      sum = (aa[cnt] + sum % m) % m;
    }
    k /= 2;
    cnt++;
  }
  cout << sum << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    solve();
    sum = cnt = 0;
    for (int i = 0; i < 36; i++) {
      aa[i] = 0;
    }
  }
}

### Prompt

In cpp, your task is to solve 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 m = 1e9 + 7;
long long n, k, aa[10001], sum, cnt;
void solve() {
  cin >> n >> k;
  aa[0] = 1;
  for (int i = 1; i <= 32; i++) {
    aa[i] = ((aa[i - 1] % m) * (n % m)) % m;
  }
  while (k) {
    if (k % 2 != 0) {
      sum = (aa[cnt] + sum % m) % m;
    }
    k /= 2;
    cnt++;
  }
  cout << sum << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    solve();
    sum = cnt = 0;
    for (int i = 0; i < 36; i++) {
      aa[i] = 0;
    }
  }
}

```