#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e3 + 10;
const int mod = 1e9 + 7;
const int base = 3;
long long sum[maxn];
string f(long long k) {
  string x = "";
  while (k) {
    x += (k % 2) + '0';
    k >>= 1;
  }
  return x;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n, k;
    scanf("%lld%lld", &n, &k);
    string tmp = f(k);
    long long ans = 0;
    long long tmpp = 1;
    for (auto ch : tmp) {
      if (ch == '1') ans += tmpp, ans %= mod;
      tmpp *= n;
      tmpp %= mod;
    }
    printf("%lld\n", ans);
  }
}

### Prompt

Your task is to create a cpp solution 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;
const int maxn = 3e3 + 10;
const int mod = 1e9 + 7;
const int base = 3;
long long sum[maxn];
string f(long long k) {
  string x = "";
  while (k) {
    x += (k % 2) + '0';
    k >>= 1;
  }
  return x;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long long n, k;
    scanf("%lld%lld", &n, &k);
    string tmp = f(k);
    long long ans = 0;
    long long tmpp = 1;
    for (auto ch : tmp) {
      if (ch == '1') ans += tmpp, ans %= mod;
      tmpp *= n;
      tmpp %= mod;
    }
    printf("%lld\n", ans);
  }
}

```