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[100005], sum, x;
int n, cnt, k;
long long mod = 1000000007ll;
multiset<long long> ms;
multiset<long long>::iterator it;
long long f(long long x, long long y) {
  if (y == 0ll)
    return 1ll;
  else if (y == 1ll)
    return x % mod;
  else {
    long long tmp = f(x, y / 2ll);
    tmp = tmp * tmp % mod;
    if (y % 2ll == 0ll)
      return tmp;
    else
      return tmp * x % mod;
  }
}
void print_ms() {
  for (it = ms.begin(); it != ms.end(); it++) printf("%lld ", *it);
  printf("\n------------\n");
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = (0); i < (n); i++) {
    scanf("%lld", &A[i]);
    sum += A[i];
  }
  for (int i = (0); i < (n); i++) A[i] = sum - A[i];
  for (int i = (0); i < (n); i++) ms.insert(A[i]);
  while (1) {
    x = *ms.begin();
    cnt = ms.count(x);
    if (cnt % k != 0)
      break;
    else {
      for (int i = (0); i < (cnt); i++) ms.erase(ms.find(x));
      cnt = cnt / k;
      for (int i = (0); i < (cnt); i++) ms.insert(x + 1ll);
    }
  }
  printf("%lld\n", f((long long)k, min(x, sum)));
  return 0;
}
