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;
void _fill_int(int* p, int val, int rep) {
  int i;
  for (i = 0; i < rep; i++) p[i] = val;
}
signed long long GETi() {
  signed long long i;
  scanf("%lld", &i);
  return i;
}
signed long long mo = 1000000007;
int N, X;
signed long long A[100001], XP[21];
signed long long modpow(signed long long a, signed long long n,
                        signed long long mo) {
  signed long long r = 1;
  while (n) {
    if (n % 2) r = (r * a) % mo;
    a = (a * a) % mo;
    n >>= 1;
  }
  return r;
}
void solve() {
  int f, i, j, k, l;
  signed long long x, y, N;
  cin >> N >> X;
  for (i = 0; i < N; i++) cin >> A[i];
  XP[0] = 1;
  for (i = 0; i < 20; i++) {
    if (XP[i] == -1) {
      XP[i + 1] = -1;
    } else {
      XP[i + 1] = XP[i] * X;
      if (XP[i + 1] > N) XP[i + 1] = -1;
    }
  }
  signed long long s = 0;
  for (i = 0; i < N; i++) s += A[i];
  for (i = 0; i < N; i++) A[i] = s - A[i];
  sort(A, A + N);
  y = A[0];
  for (i = 0; i < N; i++) A[i] -= y;
  x = 0;
  for (i = 0; i < N; i++)
    if (A[i] <= 20 && XP[A[i]] >= 0) x += XP[A[i]];
  while (x > 0 && (x % X) == 0) y++, x /= X;
  y = min(y, s);
  (void)printf("%lld\n", modpow(X, y, mo));
}
int main(int argc, char** argv) {
  if (argc > 1) freopen(argv[1], "r", stdin);
  solve();
  return 0;
}
