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 mo = 1000000007;
long long add(long long x, long long na) {
  if (na == 0) return 1;
  if (na == 1) return x % mo;
  if (na % 2 == 0)
    return (add((x * x) % mo, na / 2)) % mo;
  else
    return (add((x * x) % mo, na / 2) % mo * x) % mo;
}
int main() {
  long long a, s, d, i, j;
  cin >> a >> s;
  map<long long, long long> ma;
  map<long long, long long>::iterator ii;
  long long sum = 0;
  vector<long long> v;
  for (i = 0; i < a; i++) {
    cin >> d;
    v.push_back(d);
    sum += d;
  }
  for (i = 0; i < a; i++) {
    ma[sum - v[i]]++;
  }
  while (1) {
    ii = ma.begin();
    long long ch = ii->second;
    long long a1 = ii->first, a2 = ii->second;
    if (ch % s != 0) {
      long long otv = add(s, min(a1, sum));
      cout << otv;
      return 0;
    } else {
      ma[a1 + 1] += a2 / s;
      ma.erase(ii);
    }
  }
}
