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;
const int MOD = 1e9 + 7;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x;
}
inline int qpow(int x, long long y) {
  long long r = 1, k = x % MOD;
  for (; y; y >>= 1) {
    if (y & 1) r *= k, r %= MOD;
    k *= k, k %= MOD;
  }
  return r;
}
int main() {
  int n = read(), x = read();
  vector<long long> dat, v;
  dat.reserve(n + 1);
  long long sum = 0;
  for (int i = 1; i <= n; i++) dat[i] = read(), sum += dat[i];
  for (int i = 1; i <= n; i++) v.push_back(sum - dat[i]);
  sort(v.begin(), v.end(), greater<int>());
  while (1) {
    long long cur = v[v.size() - 1], cnt = 0;
    while (v.size() && v[v.size() - 1] == cur) cnt++, v.pop_back();
    if (cnt % x) {
      printf("%d\n", qpow(x, min(sum, cur)));
      return 0;
    } else {
      cnt /= x;
      for (int i = 1; i <= cnt; i++) v.push_back(cur + 1);
    }
  }
}
