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;
int n, a[100010];
long long sum = 0, power, x, mx = -1;
long long Pow(long long x, long long n) {
  long long y = 1, u = x % 1000000007;
  while (n) {
    if (n & 1) y = y * u % 1000000007;
    n >>= 1;
    u = u * u % 1000000007;
  }
  return y;
}
int main(int argc, char **argv) {
  scanf("%d%lld", &n, &x);
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    sum += a[i];
    mx = max(mx, (long long)a[i]);
  }
  long long tmp = sum - mx, tt = 0;
  for (int i = 0; i < n; ++i) power = (power + Pow(x, mx - a[i])) % 1000000007;
  while (power > 0 && power % x == 0) {
    ++tt;
    power /= x;
  }
  printf("%lld\n", Pow(x, min(tmp + tt, sum)));
  return 0;
}
