#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    long long ans = 0;
    vector<long long> rank(32, 0), pn(32, 0);
    rank[0] = 1, pn[0] = 1;
    for (int i = 1; i < 31; i++) {
      rank[i] = (1ll << i) + rank[i - 1];
      pn[i] = (pn[i - 1] * n) % mod;
    }
    while (k > 0) {
      int p = 0;
      for (int i = 0; i < 31; i++) {
        if (rank[i] >= k) {
          p = i;
          break;
        }
      }
      ans = (ans + pn[p]) % mod;
      k = k - rank[p - 1] - 1;
    }
    cout << ans << '\n';
  }
  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;
const int mod = 1e9 + 7;
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int T;
  cin >> T;
  while (T--) {
    int n, k;
    cin >> n >> k;
    long long ans = 0;
    vector<long long> rank(32, 0), pn(32, 0);
    rank[0] = 1, pn[0] = 1;
    for (int i = 1; i < 31; i++) {
      rank[i] = (1ll << i) + rank[i - 1];
      pn[i] = (pn[i - 1] * n) % mod;
    }
    while (k > 0) {
      int p = 0;
      for (int i = 0; i < 31; i++) {
        if (rank[i] >= k) {
          p = i;
          break;
        }
      }
      ans = (ans + pn[p]) % mod;
      k = k - rank[p - 1] - 1;
    }
    cout << ans << '\n';
  }
  return 0;
}

```