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;
int x, n;
long long a[100010];
long long Pow(long long var, long long times) {
  long long res = 1;
  while (times) {
    if (times & 1) res = res * var % 1000000007;
    var = var * var % 1000000007;
    times >>= 1LL;
  }
  return res;
}
struct cmp {
  bool operator()(const long long& x, const long long& y) { return x > y; }
};
int main() {
  scanf("%d%d", &n, &x);
  long long sum = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", a + i);
    sum += a[i];
  }
  priority_queue<long long, vector<long long>, cmp> q;
  for (int i = 1; i <= n; i++) {
    q.push(sum - a[i]);
  }
  long long t;
  while (!q.empty()) {
    t = q.top();
    int cnt = 0;
    while (!q.empty() && (q.top() == t)) {
      cnt++;
      q.pop();
      if (cnt == x) break;
    }
    if (cnt == x)
      q.push(t + 1);
    else
      break;
  }
  t = min(t, sum);
  long long ans = Pow(x, t);
  printf("%I64d\n", ans);
  return 0;
}
