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;
int n;
long long x;
long long a[100050];
long long ans;
long long sum = 0, lastsum = 0, nowsum = 0;
long long pow(long long x, long long y) {
  long long temp = 1;
  while (y) {
    if (y & 1) {
      temp *= x;
      temp %= mod;
    }
    x *= x;
    x %= mod;
    y >>= 1;
  }
  return temp % mod;
}
int main() {
  scanf("%d%lld", &n, &x);
  sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    sum += a[i];
  }
  lastsum = sum;
  sum -= a[n];
  long long now = 0;
  long long temp = a[n];
  a[0] = INT_MIN;
  for (int i = n; i >= 0; i--) {
    if (temp == a[i]) {
      now++;
    } else {
      if (now % x == 0) {
        temp = a[i];
        while ((nowsum < a[i + 1] - a[i] || i == 0) && now % x == 0) {
          now /= x;
          nowsum++;
        }
        sum += nowsum;
        if (a[i + 1] != a[i] + nowsum) break;
        now++;
        nowsum = 0;
      } else {
        break;
      }
    }
  }
  ans = pow(x, min(sum, lastsum));
  printf("%lld\n", ans);
  return 0;
}
