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 INF = 0x3f3f3f3f;
const long long MOD = 1000000007;
const int N = 1000005;
long long a[N];
long long pow_mod(long long x, long long y) {
  long long tmp = 1;
  while (y) {
    if (y & (long long)1) tmp = (tmp * x) % MOD;
    x = (x * x) % MOD;
    y = y >> 1;
  }
  return tmp;
}
int main() {
  int n, x;
  while (cin >> n >> x) {
    long long M = 0;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      M += a[i];
    }
    for (int i = 0; i < n; ++i) {
      a[i] = M - a[i];
    }
    sort(a, a + n);
    long long k1 = 0, k2 = a[0];
    for (int i = 0; i < n; ++i) {
      while (k2 < a[i] && k1 % x == 0) {
        k1 = k1 / x;
        ++k2;
      }
      if (k2 < a[i]) break;
      ++k1;
    }
    while (k1 % x == 0) {
      k1 = k1 / x;
      ++k2;
    }
    cout << pow_mod(x, min(M, k2)) << endl;
  }
  return 0;
}
