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 i, j, k, N, M, K, a[1000000];
map<long long, long long> freq;
long long big_mod(long long a, long long b, long long c) {
  if (b <= 1) {
    if (!b)
      return 1;
    else
      return a;
  } else if (!(b % 2)) {
    long long ret = big_mod(a, b / 2, c) % c;
    return (ret * ret) % c;
  } else {
    long long ret = big_mod(a, b / 2, c) % c;
    return (((ret * ret) % c) * a) % c;
  }
}
long long prime, ans, sum;
void clear() {
  sum = 0;
  ans = (long long)(1e17);
  freq.clear();
}
void add(long long x) {
  long long ind = sum - x;
  freq[ind]++;
  ans = min(ans, ind);
  if (freq[ind] == prime) {
    long long tmp = ind;
    while (freq[tmp] == prime) {
      if (ans == tmp) {
        ans++;
      }
      freq[tmp] = 0;
      freq[++tmp]++;
    }
    if (freq[tmp]) ans = min(ans, tmp);
  }
}
void init() {
  scanf("%I64d%I64d", &N, &prime);
  for (i = 1; i <= N; i++) {
    a[i] = (long long)(1e9);
    scanf("%I64d", &a[i]);
    sum += a[i];
  }
}
void calc() {
  for (i = 1; i <= N; i++) {
    add(a[i]);
  }
  ans = min(ans, sum);
  long long rb = big_mod(prime, ans, 1000000007);
  printf("%I64d\n", rb % 1000000007);
}
int main() {
  clear();
  init();
  calc();
}
