#include <bits/stdc++.h>
using namespace std;
long long int modularExponentiation(long long int x, long long int n,
                                    long long int mod) {
  long long int r = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      r = ((r % mod) * (x % mod)) % mod;
    }
    x = ((x % mod) * (x % mod)) % mod;
    n = n / 2;
  }
  return (r % mod);
}
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    long long int n, k, i;
    cin >> n >> k;
    string s = "";
    while (k > 0) {
      if (k % 2 == 1)
        s = s + '1';
      else
        s = s + '0';
      k = k / 2;
    }
    long long int ans = 0;
    for (i = 0; i < s.size(); ++i) {
      if (s[i] == '1')
        ans = (ans % 1000000007 + modularExponentiation(n, i, 1000000007)) %
              1000000007;
    }
    cout << ans % 1000000007 << endl;
  }
  return 0;
}

### Prompt

Generate 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 int modularExponentiation(long long int x, long long int n,
                                    long long int mod) {
  long long int r = 1;
  while (n > 0) {
    if (n % 2 == 1) {
      r = ((r % mod) * (x % mod)) % mod;
    }
    x = ((x % mod) * (x % mod)) % mod;
    n = n / 2;
  }
  return (r % mod);
}
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    long long int n, k, i;
    cin >> n >> k;
    string s = "";
    while (k > 0) {
      if (k % 2 == 1)
        s = s + '1';
      else
        s = s + '0';
      k = k / 2;
    }
    long long int ans = 0;
    for (i = 0; i < s.size(); ++i) {
      if (s[i] == '1')
        ans = (ans % 1000000007 + modularExponentiation(n, i, 1000000007)) %
              1000000007;
    }
    cout << ans % 1000000007 << endl;
  }
  return 0;
}

```