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;
using ll = long long;
using db = double;
using ld = long double;
using str = string;
template <class T>
using V = vector<T>;
template <class T1, class T2>
using PR = pair<T1, T2>;
template <class T1, class T2>
using umap = unordered_map<T1, T2>;
using pi = PR<int, int>;
using vi = V<int>;
const ll MOD = 1e9 + 7;
ll binexp(ll n, ll k) {
  if (k == 0) return 1;
  ll cur = binexp(n, k / 2) % MOD;
  if (k % 2)
    return ((cur * cur) % MOD * n) % MOD;
  else
    return (cur * cur) % MOD;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, x;
  cin >> n >> x;
  V<ll> a(n), b(n);
  ll sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    sum += a[i];
  }
  for (int i = 0; i < n; i++) b[i] = sum - a[i];
  while (true) {
    ll v = b[b.size() - 1];
    int occ = 0;
    while (b.size() > 0 && b[b.size() - 1] == v) {
      occ++;
      b.pop_back();
    }
    if (occ % x) {
      v = min(v, sum);
      cout << binexp((ll)x, v);
      return 0;
    } else {
      occ /= x;
      for (int i = 0; i < occ; i++) b.push_back(v + 1);
    }
  }
}
