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 arr[100000];
multiset<long long> mul;
map<long long, long long> asdf;
long long fun(long long m, int x) {
  long long a, b;
  if (asdf.find(m / 2) == asdf.end()) {
    a = fun(m / 2, x);
    asdf.insert(pair<long long, long long>(m / 2, a));
  } else
    a = asdf[m / 2];
  if (asdf.find(m - m / 2) == asdf.end()) {
    b = fun(m - m / 2, x);
    asdf.insert(pair<long long, long long>(m - m / 2, b));
  } else
    b = asdf[m - m / 2];
  return a * b % 1000000007;
}
int main() {
  int n, x;
  long long sum = 0;
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    sum += arr[i];
  }
  for (int i = 0; i < n; i++) {
    mul.insert(sum - arr[i]);
  }
  long long rst = 0;
  long long tmp = mul.count(*mul.begin());
  while (1) {
    tmp = mul.count(*mul.begin());
    if (mul.count(*mul.begin()) % x) {
      rst = *mul.begin();
      break;
    } else {
      long long qwer = *mul.begin();
      while (tmp % x == 0) {
        tmp = tmp / x;
        qwer++;
      }
      for (int i = 0; i < tmp; i++) mul.insert(qwer);
      mul.erase(*mul.begin());
    }
  }
  asdf.insert(pair<long long, long long>(0, 1));
  asdf.insert(pair<long long, long long>(1, x));
  rst = rst > sum ? sum : rst;
  cout << fun(rst, x) << endl;
  return 0;
}
