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 a[100005];
long long BigMod(long long a, long long p, long long m) {
  if (!p) return 1;
  if (p % 2) return (a % m * BigMod(a, p - 1, m) % m) % m;
  long long ta = BigMod(a, p / 2, m);
  return ((ta % m) * (ta % m)) % m;
}
int main() {
  int i, j, k;
  long long n, x;
  cin >> n >> x;
  long long tot = 0;
  long long mx = 0;
  for (i = 0; i < (n); i++) {
    scanf("%I64d", &a[i]);
    tot += a[i];
  }
  map<long long, int> Map;
  for (i = 0; i < (n); i++) {
    Map[tot - a[i]]++;
  }
  priority_queue<long long> pq;
  long long u, v;
  for (map<long long, int>::iterator it = Map.begin(); it != Map.end(); it++) {
    pq.push(-(it->first));
  }
  while (true) {
    u = -pq.top();
    pq.pop();
    v = Map[u];
    Map.erase(Map.find(u));
    if (v % x) break;
    while (v % x == 0) {
      v /= x;
      u++;
    }
    if (Map.find(u) == Map.end()) {
      pq.push(-u);
      Map[u] = v;
    } else
      Map[u] += v;
  }
  u = min(u, tot);
  long long ans = BigMod(x, u, 1000000007);
  cout << ans << endl;
  return 0;
}
