#include <bits/stdc++.h>
using namespace std;
long long bigmod(long long b, long long p) {
  if (p == 0) return 1;
  if (p == 1) return b;
  long long a = bigmod(b, p / 2);
  a *= a;
  a %= 1000000007;
  if (p % 2 == 1) a *= b;
  a %= 1000000007;
  return a;
}
signed main() {
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long p = 1;
    long long ans = 0;
    while (k > 0) {
      ans += p * (k % 2);
      k /= 2;
      p *= n;
      p %= 1000000007;
      ans %= 1000000007;
    }
    cout << ans << "\n";
  }
}

### 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;
long long bigmod(long long b, long long p) {
  if (p == 0) return 1;
  if (p == 1) return b;
  long long a = bigmod(b, p / 2);
  a *= a;
  a %= 1000000007;
  if (p % 2 == 1) a *= b;
  a %= 1000000007;
  return a;
}
signed main() {
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long p = 1;
    long long ans = 0;
    while (k > 0) {
      ans += p * (k % 2);
      k /= 2;
      p *= n;
      p %= 1000000007;
      ans %= 1000000007;
    }
    cout << ans << "\n";
  }
}

```