#include <bits/stdc++.h>
using namespace ::std;
long long int MOD = 1000000007;
long long int a[100001], b[100001];
long long int power(long long int base, long long int exp1) {
  if (exp1 == 0) return 1;
  long long int val1 = power(base, exp1 / 2);
  long long int ans = (val1 * val1) % MOD;
  if (exp1 & 1) ans = (ans * base) % MOD;
  return ans;
}
long long int solveRecursively(long long int n, long long int k,
                               long long int numBits) {
  if (k == 0) return 0;
  if (numBits == 1) {
    if (k == 1)
      return 1;
    else
      return 0;
  } else if (double(k) > (double(power(2, numBits - 1) - 1))) {
    return ((((long long int)power(n, numBits - 1)) % MOD) +
            (solveRecursively(n, k - power(2, numBits - 1), numBits - 1) %
             MOD)) %
           MOD;
  } else {
    return solveRecursively(n, k, numBits - 1) % MOD;
  }
}
long long int solve(long long int n, long long int k) {
  long long int interm = k;
  long long int num = 0;
  while (interm > 0) {
    num++;
    interm = interm / 2;
  }
  return solveRecursively(n, k, num);
}
vector<long long int> v;
int main() {
  long long int t, n, m, m2, j, k, i, x;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    cout << solve(n, k) << endl;
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 int MOD = 1000000007;
long long int a[100001], b[100001];
long long int power(long long int base, long long int exp1) {
  if (exp1 == 0) return 1;
  long long int val1 = power(base, exp1 / 2);
  long long int ans = (val1 * val1) % MOD;
  if (exp1 & 1) ans = (ans * base) % MOD;
  return ans;
}
long long int solveRecursively(long long int n, long long int k,
                               long long int numBits) {
  if (k == 0) return 0;
  if (numBits == 1) {
    if (k == 1)
      return 1;
    else
      return 0;
  } else if (double(k) > (double(power(2, numBits - 1) - 1))) {
    return ((((long long int)power(n, numBits - 1)) % MOD) +
            (solveRecursively(n, k - power(2, numBits - 1), numBits - 1) %
             MOD)) %
           MOD;
  } else {
    return solveRecursively(n, k, numBits - 1) % MOD;
  }
}
long long int solve(long long int n, long long int k) {
  long long int interm = k;
  long long int num = 0;
  while (interm > 0) {
    num++;
    interm = interm / 2;
  }
  return solveRecursively(n, k, num);
}
vector<long long int> v;
int main() {
  long long int t, n, m, m2, j, k, i, x;
  cin >> t;
  while (t--) {
    cin >> n >> k;
    cout << solve(n, k) << endl;
  }
  return 0;
}

```