#include <bits/stdc++.h>
using namespace std;
long long int N = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, k, i;
    cin >> n >> k;
    vector<int> v;
    v.push_back(1);
    for (i = 1; i < 30; i++) {
      v.push_back(((v[i - 1] % N) * (n % N)) % N);
      if (v[i] < 0) v[i] += N;
    }
    int binaryNum[30], b[30];
    for (i = 0; i < 30; i++) binaryNum[i] = 0;
    i = 0;
    while (k > 0) {
      binaryNum[i] = k % 2;
      k = k / 2;
      i++;
    }
    int ans = 0;
    for (i = 0; i < 30; i++) {
      ans = ((ans % N) + (binaryNum[i] * v[i]) % N) % N;
      if (ans < 0) ans += N;
    }
    cout << ans << endl;
  }
}

### Prompt

Your challenge is to write 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 N = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n, k, i;
    cin >> n >> k;
    vector<int> v;
    v.push_back(1);
    for (i = 1; i < 30; i++) {
      v.push_back(((v[i - 1] % N) * (n % N)) % N);
      if (v[i] < 0) v[i] += N;
    }
    int binaryNum[30], b[30];
    for (i = 0; i < 30; i++) binaryNum[i] = 0;
    i = 0;
    while (k > 0) {
      binaryNum[i] = k % 2;
      k = k / 2;
      i++;
    }
    int ans = 0;
    for (i = 0; i < 30; i++) {
      ans = ((ans % N) + (binaryNum[i] * v[i]) % N) % N;
      if (ans < 0) ans += N;
    }
    cout << ans << endl;
  }
}

```