#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long mod = 1e9 + 7;
const long long base = 31;
long long power(long long a, long long n, long long mod) {
  if (n == 0)
    return 1;
  else {
    long long temp = power(a, n / 2, mod) % mod;
    temp = (temp * temp) % mod;
    if (n % 2 == 1)
      return (a * temp) % mod;
    else
      return temp;
  }
}
void fre() {}
clock_t start = clock();
double timecomplexity;
int n, k;
long long ans = 0;
void solve() {
  cin >> n >> k;
  ans = 0;
  for (int i = 0; i <= 31; i++)
    if (((k >> i) & 1) == 1) {
      ans += power(n, i, mod);
      ans %= mod;
    }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>
#pragma GCC optimize("O3")
using namespace std;
const long long mod = 1e9 + 7;
const long long base = 31;
long long power(long long a, long long n, long long mod) {
  if (n == 0)
    return 1;
  else {
    long long temp = power(a, n / 2, mod) % mod;
    temp = (temp * temp) % mod;
    if (n % 2 == 1)
      return (a * temp) % mod;
    else
      return temp;
  }
}
void fre() {}
clock_t start = clock();
double timecomplexity;
int n, k;
long long ans = 0;
void solve() {
  cin >> n >> k;
  ans = 0;
  for (int i = 0; i <= 31; i++)
    if (((k >> i) & 1) == 1) {
      ans += power(n, i, mod);
      ans %= mod;
    }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}

```