#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
long long gcd(long long a, long long b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long binopow(long long b, long long pow) {
  if (pow == 0) return 1;
  long long ans;
  ans = binopow(b, pow / 2);
  ans = (ans * ans) % mod;
  if (pow % 2) ans = (ans * b) % mod;
  return ans;
}
long long inv(long long a, long long mod) { return binopow(a, mod - 2); }
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<int> v;
    while (k > 0) {
      v.push_back(k % 2);
      k = k / 2;
    }
    long long ans = 0;
    long long j = 1;
    for (int i = 0; i < v.size(); i++) {
      ans = ans + (v[i] * j) % mod;
      ans = ans % mod;
      j = (j * n) % mod;
    }
    if (ans < 0)
      cout << ans + mod;
    else
      cout << ans << "\n";
  }
}

### Prompt

Please formulate 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;
long long gcd(long long a, long long b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long binopow(long long b, long long pow) {
  if (pow == 0) return 1;
  long long ans;
  ans = binopow(b, pow / 2);
  ans = (ans * ans) % mod;
  if (pow % 2) ans = (ans * b) % mod;
  return ans;
}
long long inv(long long a, long long mod) { return binopow(a, mod - 2); }
int main() {
  int t;
  cin >> t;
  while (t--) {
    long long n, k;
    cin >> n >> k;
    vector<int> v;
    while (k > 0) {
      v.push_back(k % 2);
      k = k / 2;
    }
    long long ans = 0;
    long long j = 1;
    for (int i = 0; i < v.size(); i++) {
      ans = ans + (v[i] * j) % mod;
      ans = ans % mod;
      j = (j * n) % mod;
    }
    if (ans < 0)
      cout << ans + mod;
    else
      cout << ans << "\n";
  }
}

```