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;
const long long mod = 1000000007;
long long a[111111];
long long n, x;
long long cal(long long x, long long k) {
  if (k == 0) return 1;
  if (k == 1) return x;
  long long ret = cal(x, k / 2);
  if (k & 1) return ((ret * ret) % mod * x) % mod;
  return (ret * ret) % mod;
}
map<long long, int> M;
int main() {
  while (scanf("%I64d%I64d", &n, &x) == 2) {
    M.clear();
    long long sum = 0;
    for (int i = 0; i < n; ++i) {
      scanf("%I64d", &a[i]);
      sum += a[i];
    }
    for (int i = 0; i < n; ++i) M[a[n - 1] - a[i]]++;
    long long p = sum - a[n - 1];
    for (map<long long, int>::iterator it = M.begin(); it != M.end(); ++it) {
      long long p1 = it->first, p2 = it->second, p3 = it->first;
      if (p2 % x) {
        p += p1;
        break;
      }
      while (p2 % x == 0) {
        p2 /= x;
        p1++;
      }
      M[p1] += p2;
    }
    long long ans = cal(x, min(sum, p));
    cout << ans << "\n";
  }
  return 0;
}
