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 a[100666];
map<long long, int> fr;
long long i, x, st, numar;
long long n;
long long lgpow(long long val, long long P) {
  if (P == 1) return val;
  if (P == 0) return 1;
  long long V = lgpow(val, P / 2);
  V *= V;
  V %= 1000000007;
  if (P % 2) V *= val;
  V %= 1000000007;
  return V;
}
int main() {
  cin >> n >> x;
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    st += a[i];
  }
  long long vmin = ((long long)1 << (long long)60);
  for (i = 1; i <= n; ++i) {
    long long V = st - a[i];
    fr[V]++;
    vmin = min(vmin, V);
  }
  for (i = vmin; i <= st; ++i) {
    if (fr.find(i) != fr.end()) numar += fr[i];
    if (numar % x) {
      cout << lgpow(x, i);
      return 0;
    }
    long long h = numar / x;
    fr[i + 1] += h;
  }
  cout << lgpow(x, st);
  return 0;
}
