#include <bits/stdc++.h>
const long long mod = 1000000000 + 7;
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fastPower(long long a, long long b) {
  if (b == 0) return 1;
  long long value = fastPower(a, b / 2);
  if (b % 2 == 0)
    return value * value;
  else
    return value * value * a;
}
long long fastPowerMod(long long x, long long y, long long mod) {
  if (y == 0) return 1 % mod;
  long long u = fastPowerMod(x, y / 2, mod);
  u = u * u % mod;
  if (y & 1) u = u * x % mod;
  return u;
}
int main() {
  long long t, n, m;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    long long com, f, ans = 0;
    do {
      com = (long long)log2(m);
      f = fastPowerMod(2, com, mod);
      ans += fastPowerMod(n, com, mod);
      m -= f;
    } while (m != 0);
    cout << ans % mod << "\n";
  }
}

### Prompt

Generate 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>
const long long mod = 1000000000 + 7;
using namespace std;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fastPower(long long a, long long b) {
  if (b == 0) return 1;
  long long value = fastPower(a, b / 2);
  if (b % 2 == 0)
    return value * value;
  else
    return value * value * a;
}
long long fastPowerMod(long long x, long long y, long long mod) {
  if (y == 0) return 1 % mod;
  long long u = fastPowerMod(x, y / 2, mod);
  u = u * u % mod;
  if (y & 1) u = u * x % mod;
  return u;
}
int main() {
  long long t, n, m;
  cin >> t;
  while (t--) {
    cin >> n >> m;
    long long com, f, ans = 0;
    do {
      com = (long long)log2(m);
      f = fastPowerMod(2, com, mod);
      ans += fastPowerMod(n, com, mod);
      m -= f;
    } while (m != 0);
    cout << ans % mod << "\n";
  }
}

```