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 Maxn = 1e5 + 19, Mod = 1e9 + 7;
vector<long long> S;
long long sum, A[Maxn], n, x, now;
inline long long Power(long long a, long long b) {
  long long Res = 1, tmp = a;
  while (b) {
    if (b & 1) Res = Res * tmp % Mod;
    tmp = tmp * tmp % Mod;
    b >>= 1;
  }
  return Res;
}
int main() {
  scanf("%d%d", &n, &x);
  for (int i = 0; i < n; i++) scanf("%I64d", &A[i]), sum += A[i];
  for (int i = 0; i < n; i++) S.push_back(sum - A[i]);
  while (1) {
    long long v = S[S.size() - 1], cnt = 0;
    while (S.size() && S[S.size() - 1] == v) cnt++, S.pop_back();
    if (cnt % x == 0) {
      cnt /= x;
      for (int i = 0; i < cnt; i++) S.push_back(v + 1);
    } else {
      printf("%I64d\n", Power(x, min(1LL * v, sum)));
      return 0;
    }
  }
}
