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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    long long n, x;
    cin >> n >> x;
    vector<long long> a(n);
    long long s = 0;
    long long sm = 0;
    for (long long i = 0; i < n; i++) cin >> a[i], s += a[i], sm += a[i];
    for (long long i = 0; i < n; i++) a[i] = s - a[i];
    reverse(a.begin(), a.end());
    long long flag = 1;
    s = a[0];
    for (long long i = 0; i < n; i++) a[i] -= s;
    long long add = 0;
    while (flag) {
      vector<long long> v;
      long long cnt0 = 0;
      for (long long i = 0; i < a.size(); i++) {
        if (a[i] == 0)
          cnt0++;
        else {
          if (cnt0 % x != 0) {
            flag = 0;
            break;
          } else {
            v.push_back(a[i] - 1);
            cnt0 = 0;
          }
        }
        if (cnt0 == x) {
          v.push_back(0);
          cnt0 = 0;
        }
      }
      if (cnt0 < x and cnt0 > 0) {
        flag = 0;
      }
      if (flag) {
        a.clear();
        s++;
        a = v;
      }
    }
    long long result = 1;
    if (s > sm) s = sm;
    while (s) {
      if (s % 2)
        result = (result * x) % 1000000007, s--;
      else {
        x = (x * x) % 1000000007;
        s /= 2;
      }
    }
    cout << result;
  }
}
