#include <bits/stdc++.h>
using namespace std;
long long rem = 1000000007;
long long binaryexpo(long long x, long long n, long long m) {
  long long res = 1;
  while (n != 0) {
    if (n % 2 == 1) {
      res = ((res % m) * (x % m)) % m;
    }
    x = ((x % m) * (x % m)) % m;
    n /= 2;
  }
  return res;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long temp = log2(k);
    long long terms = pow(2, temp);
    if (k == terms) {
      long long var = binaryexpo(n, temp, rem);
      cout << var << "\n";
    } else {
      long long left = k - terms;
      long long ans = binaryexpo(n, temp, rem);
      while (left > 0) {
        long long go = log2(left);
        long long ter = pow(2, go);
        left -= ter;
        long long ad = binaryexpo(n, go, rem);
        ans = (ans + ad) % rem;
      }
      cout << ans << "\n";
    }
  }
}

### Prompt

Please formulate a Cpp solution to 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 rem = 1000000007;
long long binaryexpo(long long x, long long n, long long m) {
  long long res = 1;
  while (n != 0) {
    if (n % 2 == 1) {
      res = ((res % m) * (x % m)) % m;
    }
    x = ((x % m) * (x % m)) % m;
    n /= 2;
  }
  return res;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long temp = log2(k);
    long long terms = pow(2, temp);
    if (k == terms) {
      long long var = binaryexpo(n, temp, rem);
      cout << var << "\n";
    } else {
      long long left = k - terms;
      long long ans = binaryexpo(n, temp, rem);
      while (left > 0) {
        long long go = log2(left);
        long long ter = pow(2, go);
        left -= ter;
        long long ad = binaryexpo(n, go, rem);
        ans = (ans + ad) % rem;
      }
      cout << ans << "\n";
    }
  }
}

```