Simon has a prime number x and an array of non-negative integers a1, a2, ..., an.

Simon loves fractions very much. Today he wrote out number <image> on a piece of paper. After Simon led all fractions to a common denominator and summed them up, he got a fraction: <image>, where number t equals xa1 + a2 + ... + an. Now Simon wants to reduce the resulting fraction. 

Help him, find the greatest common divisor of numbers s and t. As GCD can be rather large, print it as a remainder after dividing it by number 1000000007 (109 + 7).

Input

The first line contains two positive integers n and x (1 ≤ n ≤ 105, 2 ≤ x ≤ 109) — the size of the array and the prime number.

The second line contains n space-separated integers a1, a2, ..., an (0 ≤ a1 ≤ a2 ≤ ... ≤ an ≤ 109). 

Output

Print a single number — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

2 2
2 2


Output

8


Input

3 3
1 2 3


Output

27


Input

2 2
29 29


Output

73741817


Input

4 5
0 0 0 0


Output

1

Note

In the first sample <image>. Thus, the answer to the problem is 8.

In the second sample, <image>. The answer to the problem is 27, as 351 = 13·27, 729 = 27·27.

In the third sample the answer to the problem is 1073741824 mod 1000000007 = 73741817.

In the fourth sample <image>. Thus, the answer to the problem is 1.
#include <bits/stdc++.h>
using namespace std;
long long powmod(long long x, long long n) {
  if (n == 0) return 1;
  if (n % 2)
    return (x * powmod((x * x) % (((long long)1e9) + 7), n / 2)) %
           (((long long)1e9) + 7);
  return powmod((x * x) % (((long long)1e9) + 7), n / 2) %
         (((long long)1e9) + 7);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, x;
  cin >> n >> x;
  long long a[n];
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  priority_queue<pair<long long, long long> > h;
  for (int i = 0; i < n; i++) h.push(make_pair(a[i] - sum, 1));
  long long k = 0;
  long long mp = -h.top().first;
  while (h.size() > 1) {
    long long f = h.top().second;
    h.pop();
    while (h.size()) {
      long long ap = -h.top().first;
      if (ap != mp) break;
      f += h.top().second;
      h.pop();
    }
    if (f % x) break;
    int po = 0;
    while (f % x == 0) {
      po++;
      f /= x;
    }
    h.push(make_pair(-(mp + po), f));
    mp = -h.top().first;
  }
  cout << powmod(x, min(sum, mp));
  return 0;
}
