#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
vector<long long int> arr(62);
void calc() {
  arr[0] = 1;
  for (int i = 1; i < 62; i++) {
    arr[i] = arr[i - 1] * 2;
  }
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2) ans = (ans * a) % mod;
    b /= 2;
    a = (a * a) % mod;
  }
  return ans;
}
void test_case() {
  long long int n, k;
  cin >> n >> k;
  long long int ans = 0;
  while (k > 0) {
    int index = lower_bound(arr.begin(), arr.end(), k) - arr.begin();
    if (arr[index] != k) index--;
    k -= arr[index];
    ans += power(n, index);
    ans %= mod;
  }
  cout << ans << endl;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    calc();
    test_case();
  }
}

### 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 long long int mod = 1e9 + 7;
vector<long long int> arr(62);
void calc() {
  arr[0] = 1;
  for (int i = 1; i < 62; i++) {
    arr[i] = arr[i - 1] * 2;
  }
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b % 2) ans = (ans * a) % mod;
    b /= 2;
    a = (a * a) % mod;
  }
  return ans;
}
void test_case() {
  long long int n, k;
  cin >> n >> k;
  long long int ans = 0;
  while (k > 0) {
    int index = lower_bound(arr.begin(), arr.end(), k) - arr.begin();
    if (arr[index] != k) index--;
    k -= arr[index];
    ans += power(n, index);
    ans %= mod;
  }
  cout << ans << endl;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    calc();
    test_case();
  }
}

```