#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7, n, m, ans, y, cnt;
string s;
short j;
int inn() {
  int x;
  cin >> x;
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int _tc;
  cin >> _tc;
  while (_tc--) {
    n = inn();
    m = inn();
    ans = 0;
    j = 0;
    s = bitset<64>(m).to_string();
    reverse(s.begin(), s.end());
    long long h = 1;
    for (int i = 0; i < s.size(); i++) {
      h *= n;
      h %= mod;
      if (j == 0) h = 1;
      if (s[i] == '1') {
        ans += h % mod;
        ans %= mod;
      }
      j++;
    }
    cout << ans << "\n";
  }
  return 0;
}

### 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;
long long mod = 1e9 + 7, n, m, ans, y, cnt;
string s;
short j;
int inn() {
  int x;
  cin >> x;
  return x;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int _tc;
  cin >> _tc;
  while (_tc--) {
    n = inn();
    m = inn();
    ans = 0;
    j = 0;
    s = bitset<64>(m).to_string();
    reverse(s.begin(), s.end());
    long long h = 1;
    for (int i = 0; i < s.size(); i++) {
      h *= n;
      h %= mod;
      if (j == 0) h = 1;
      if (s[i] == '1') {
        ans += h % mod;
        ans %= mod;
      }
      j++;
    }
    cout << ans << "\n";
  }
  return 0;
}

```