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 FASTBUFFER;
multiset<long long> b;
multiset<long long>::iterator it;
long long n, x, i, j, Ans, Min, Now, Sur, s, a[100005];
long long sumi(long long a, long long b);
int main() {
  scanf("%I64d %I64d", &n, &x);
  for (i = 1; i <= n; i++) scanf("%I64d", &a[i]), s += a[i];
  for (i = 1; i <= n; i++) b.insert(s - a[i]);
  Now = *b.begin(), Sur = 0, it = b.begin();
  while (1) {
    for (; it != b.end(); ++it)
      if (*it == Now)
        Sur++;
      else
        break;
    if (Sur % x > 0) {
      Min = Now;
      break;
    }
    Sur /= x;
    Now++;
  }
  if (Min > s) Min = s;
  Ans = sumi(x, Min);
  printf("%I64d\n", Ans);
  return 0;
}
long long sumi(long long a, long long b) {
  long long ret = 1, c = a;
  while (b > 0) {
    if (b & 1LL) ret *= c, ret %= 1000000007;
    c *= c, c %= 1000000007;
    b >>= 1;
  }
  return ret;
}
