#include <bits/stdc++.h>
const long long modulo = 1e9 + 7;
using namespace std;
long long binaryExponentiation(long long x, long long n, long long m) {
  if (n == 0) {
    return 1;
  }
  if (n % 2 == 0) {
    return binaryExponentiation(((x % m) * (x % m)) % m, n / 2, m);
  }
  return ((x % m) *
          binaryExponentiation(((x % m) * (x % m)) % m, (n - 1) / 2, m) % m) %
         m;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long i = 0, j = 1;
    while (k / j > 0) {
      j *= 2;
      i++;
    }
    long long s[i];
    for (long long p = 0; p < i; p++) {
      if (((1LL << p) & k)) {
        s[p] = 1;
      } else {
        s[p] = 0;
      }
    }
    long long ans = 0;
    for (long long p = 0; p < i; p++) {
      if (s[p] == 1) {
        ans += binaryExponentiation(n, p, modulo);
        ans %= modulo;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### 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>
const long long modulo = 1e9 + 7;
using namespace std;
long long binaryExponentiation(long long x, long long n, long long m) {
  if (n == 0) {
    return 1;
  }
  if (n % 2 == 0) {
    return binaryExponentiation(((x % m) * (x % m)) % m, n / 2, m);
  }
  return ((x % m) *
          binaryExponentiation(((x % m) * (x % m)) % m, (n - 1) / 2, m) % m) %
         m;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    long long i = 0, j = 1;
    while (k / j > 0) {
      j *= 2;
      i++;
    }
    long long s[i];
    for (long long p = 0; p < i; p++) {
      if (((1LL << p) & k)) {
        s[p] = 1;
      } else {
        s[p] = 0;
      }
    }
    long long ans = 0;
    for (long long p = 0; p < i; p++) {
      if (s[p] == 1) {
        ans += binaryExponentiation(n, p, modulo);
        ans %= modulo;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```