#include <bits/stdc++.h>
using namespace std;
int64_t mod = 1e9 + 7;
int64_t log2(int64_t x, int64_t base) { return log(x) / log(base); }
int64_t bin_pow(int64_t base, int64_t pow) {
  if (pow == 0) {
    return 1;
  }
  if (pow == 1) {
    return base;
  }
  int64_t res = bin_pow(base, pow / 2);
  res %= mod;
  res *= res;
  res %= mod;
  if (pow % 2 != 0) {
    res *= base;
    res %= mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int64_t t;
  cin >> t;
  while (t--) {
    int64_t n, k;
    cin >> n >> k;
    int64_t res = 0;
    while (k > 0) {
      int64_t n2p = log2(k, 2);
      res += bin_pow(n, n2p);
      res %= mod;
      k -= pow(2, n2p);
    }
    cout << res << "\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;
int64_t mod = 1e9 + 7;
int64_t log2(int64_t x, int64_t base) { return log(x) / log(base); }
int64_t bin_pow(int64_t base, int64_t pow) {
  if (pow == 0) {
    return 1;
  }
  if (pow == 1) {
    return base;
  }
  int64_t res = bin_pow(base, pow / 2);
  res %= mod;
  res *= res;
  res %= mod;
  if (pow % 2 != 0) {
    res *= base;
    res %= mod;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int64_t t;
  cin >> t;
  while (t--) {
    int64_t n, k;
    cin >> n >> k;
    int64_t res = 0;
    while (k > 0) {
      int64_t n2p = log2(k, 2);
      res += bin_pow(n, n2p);
      res %= mod;
      k -= pow(2, n2p);
    }
    cout << res << "\n";
  }
}

```