#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
long long int binmod(long long int k, long long int x) {
  if (x == 0) return 1;
  if (x == 1) return k % mod;
  long long int res = binmod(k, x / 2);
  if (x % 2) return ((res * res % mod) * k) % mod;
  return res * res % mod;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int TESTCASE;
  cin >> TESTCASE;
  while (TESTCASE--) {
    long long int n, k;
    cin >> n >> k;
    vector<long long int> bit(32);
    for (long long int i = 0; i < 32; i++) {
      if (k & (1 << i)) {
        bit[i] = 1;
      }
    }
    long long int ans = 0;
    for (long long int i = 0; i < 32; i++) {
      if (bit[i]) ans = (ans + binmod(n, i)) % mod;
    }
    cout << ans << "\n";
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 long long int mod = 1e9 + 7;
long long int binmod(long long int k, long long int x) {
  if (x == 0) return 1;
  if (x == 1) return k % mod;
  long long int res = binmod(k, x / 2);
  if (x % 2) return ((res * res % mod) * k) % mod;
  return res * res % mod;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int TESTCASE;
  cin >> TESTCASE;
  while (TESTCASE--) {
    long long int n, k;
    cin >> n >> k;
    vector<long long int> bit(32);
    for (long long int i = 0; i < 32; i++) {
      if (k & (1 << i)) {
        bit[i] = 1;
      }
    }
    long long int ans = 0;
    for (long long int i = 0; i < 32; i++) {
      if (bit[i]) ans = (ans + binmod(n, i)) % mod;
    }
    cout << ans << "\n";
  }
  return 0;
}

```