#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int M = 2e3 + 5;
const int moh = 1e9 + 3;
const long long mod = 1e9 + 7;
const int base = 311;
const int oi = 2e9;
const long long oo = 1e18;
const double eps = 1e-9;
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[] = {-1, 0, 1, 0, -1, 1, -1, 1};
int n, k;
long long po[N];
void solve() {
  cin >> n >> k;
  po[0] = 1;
  for (int i = 1; i <= 30; i++) {
    po[i] = po[i - 1] * n % mod;
  }
  long long ans = 0;
  for (int i = 30; i >= 0; i--) {
    if (k >> i & 1) {
      ans += po[i];
      ans %= mod;
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test;
  cin >> test;
  while (test--) {
    solve();
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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 = 1e5 + 5;
const int M = 2e3 + 5;
const int moh = 1e9 + 3;
const long long mod = 1e9 + 7;
const int base = 311;
const int oi = 2e9;
const long long oo = 1e18;
const double eps = 1e-9;
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1};
int dy[] = {-1, 0, 1, 0, -1, 1, -1, 1};
int n, k;
long long po[N];
void solve() {
  cin >> n >> k;
  po[0] = 1;
  for (int i = 1; i <= 30; i++) {
    po[i] = po[i - 1] * n % mod;
  }
  long long ans = 0;
  for (int i = 30; i >= 0; i--) {
    if (k >> i & 1) {
      ans += po[i];
      ans %= mod;
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test;
  cin >> test;
  while (test--) {
    solve();
  }
}

```