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 int MAX = 1e9 + 7;
long long bpow(long long x, long long v) {
  if (v == 0) return 1 % MAX;
  if (v % 2 == 1) return (x * bpow(x, v - 1)) % MAX;
  long long r = bpow(x, v / 2);
  return (r * r) % MAX;
}
int main(void) {
  int n, x;
  cin >> n >> x;
  vector<int> a(n);
  vector<long long> sums(n);
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 0; i < n; i++) {
    sums[i] = sum - a[i];
  }
  int last = n - 1;
  while (true) {
    long long maxS = sums[last];
    int count = 0;
    while (last >= 0 && maxS == sums[last]) {
      count++;
      last--;
    }
    if (count % x != 0) {
      cout << bpow(x, min(maxS, sum));
      return 0;
    } else {
      count /= x;
      for (int i = 0; i < count; i++) {
        last++;
        sums[last] = maxS + 1;
      }
    }
  }
  return 0;
}
