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;
#pragma warning(disable : 4996)
int n, x;
int d[100010];
long long e[100010], sum, r;
int power() {
  long long res = 1, pre = x;
  while (r) {
    if (r % 2 == 1) res *= pre;
    res %= 1000000007;
    pre *= pre;
    pre %= 1000000007;
    r /= 2;
  }
  return (int)res;
}
int main() {
  scanf("%d%d", &n, &x);
  for (int i = 0; i < n; i++) {
    scanf("%d", &d[i]);
    sum += d[i];
  };
  for (int i = 0; i < n; i++) e[i] = sum - d[i];
  sort(e, e + n);
  int p1 = 0, p2 = 0;
  e[n] = -1;
  while (p1 < n) {
    while (e[p1] == e[p2]) p2++;
    p2--;
    if ((p2 - p1 + 1) % x != 0) {
      r = e[p1];
      break;
    }
    for (int i = 0; i < (p2 - p1 + 1) / x; i++) e[p2 - i] = e[p1] + 1;
    p1 = p2 - (p2 - p1 + 1) / x + 1;
    p2 = p1;
  }
  r = min(r, sum);
  cout << power() << endl;
  return 0;
};
