#include <bits/stdc++.h>
using namespace std;
const int N = 4e5 + 5;
int MOD = 1e9 + 7;
long long fast_power(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power % 2 == 1) {
      result = (result * base) % MOD;
    }
    base = (base * base) % MOD;
    power = power / 2;
  }
  return result;
}
int n, k;
bool get(int index) {
  long long get = fast_power(2, index);
  if (k < get) return 0;
  if (get == 1) return k % 2;
  return ((k / get) % 2);
}
int main() {
  int test;
  cin >> test;
  while (test--) {
    cin >> n >> k;
    long long sum = 0;
    for (int i = 0; i <= 30; ++i) {
      if ((k >> i) & 1) {
        sum = (sum % MOD + fast_power(n, i) % MOD) % MOD;
      }
    }
    cout << sum % MOD << "\n";
  }
  return 0;
}

### 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;
const int N = 4e5 + 5;
int MOD = 1e9 + 7;
long long fast_power(long long base, long long power) {
  long long result = 1;
  while (power > 0) {
    if (power % 2 == 1) {
      result = (result * base) % MOD;
    }
    base = (base * base) % MOD;
    power = power / 2;
  }
  return result;
}
int n, k;
bool get(int index) {
  long long get = fast_power(2, index);
  if (k < get) return 0;
  if (get == 1) return k % 2;
  return ((k / get) % 2);
}
int main() {
  int test;
  cin >> test;
  while (test--) {
    cin >> n >> k;
    long long sum = 0;
    for (int i = 0; i <= 30; ++i) {
      if ((k >> i) & 1) {
        sum = (sum % MOD + fast_power(n, i) % MOD) % MOD;
      }
    }
    cout << sum % MOD << "\n";
  }
  return 0;
}

```