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 n, x, a[1000000], p[1000000], i, j, as, s, mn, q, k, mod = 1000000007;
long long mul(long long a, long long b) {
  long long q = (long long)((long double)a * (long double)b / (long double)mod);
  long long r = a * b - q * mod;
  return (r + 5 * mod) % mod;
}
long long binpow(long long a, long long n) {
  long long res = 1;
  while (n)
    if (n & 1) {
      res = mul(res, a);
      --n;
    } else {
      a = mul(a, a);
      n >>= 1;
    }
  return res;
}
int main() {
  scanf("%I64d%I64d", &n, &x);
  map<long long, long long> m;
  as = 0;
  for (i = 0; i < n; i++) scanf("%I64d", &a[i]), as += a[i];
  for (i = 0; i < n; i++) {
    p[i] = as - a[i];
    m[p[i]]++;
  }
  for (map<long long, long long>::iterator it = m.begin(); it != m.end();
       it++) {
    if (it->second / x) m[it->first + 1] += it->second / x;
    it->second %= x;
  }
  for (map<long long, long long>::iterator it = m.begin(); it != m.end(); it++)
    if (it->second) {
      cout << binpow(x, min(it->first, as));
      return 0;
    }
  cout << 1;
  return 0;
}
