#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
int64_t t, n, k;
int64_t power(int64_t a, int64_t b) {
  int64_t ans = 1;
  for (int i = 1; i <= b; i++) {
    ans = a % mod * ans % mod;
  }
  return ans % mod;
}
void solve() {
  string s = bitset<64>(k).to_string();
  reverse(s.begin(), s.end());
  int64_t ans = 0;
  for (int64_t i = 0; i < s.size(); i++) {
    if (s[i] == '1') ans = (ans % mod + power(n, i) % mod) % mod;
  }
  cout << ans % mod << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  cin >> t;
  while (t--) {
    cin >> n >> k;
    solve();
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int64_t t, n, k;
int64_t power(int64_t a, int64_t b) {
  int64_t ans = 1;
  for (int i = 1; i <= b; i++) {
    ans = a % mod * ans % mod;
  }
  return ans % mod;
}
void solve() {
  string s = bitset<64>(k).to_string();
  reverse(s.begin(), s.end());
  int64_t ans = 0;
  for (int64_t i = 0; i < s.size(); i++) {
    if (s[i] == '1') ans = (ans % mod + power(n, i) % mod) % mod;
  }
  cout << ans % mod << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  cin >> t;
  while (t--) {
    cin >> n >> k;
    solve();
  }
  return 0;
}

```