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>
const long long INF = 2000000000000000;
const double PI = 3.14159265358979323846;
const long long MOD = 1000000000 + 7;
using namespace std;
long long n, x;
long long st[100005], t[100005], sum;
long long mint;
long long bin_pow(long long a, long long v) {
  if (v == 0) return 1;
  if (v % 2 == 1) {
    return (a * bin_pow(a, v - 1)) % MOD;
  } else {
    long long t = bin_pow(a, v / 2) % MOD;
    return t * t % MOD;
  }
}
int main() {
  mint = INF;
  cin >> n >> x;
  bool p = 0;
  for (int i(0); i < n; i++) {
    cin >> t[i];
    sum += t[i];
    if (t[i] == 0) p = 1;
  }
  for (int i(0); i < n; i++) {
    st[i] = sum - t[i];
  }
  sort(st, st + n);
  int k = 1;
  for (int i(0); i < n; i++) {
    if (i != n - 1 && st[i] == st[i + 1]) {
      k++;
    } else {
      if (k % x != 0) {
        mint = st[i];
        break;
      } else {
        k /= x;
        while (k > 0) {
          st[i]++;
          i--;
          k--;
        }
        k = 1;
      }
    }
  }
  if (mint >= sum) {
    mint = sum;
  }
  cout << bin_pow(x, mint);
}
