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 inf = (int)1e9;
const int mod = (int)1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-9;
int n;
long long x;
long long a[100500];
map<long long, long long> st;
set<long long> second;
long long first(long long a, long long k) {
  if (k == 0) return 1;
  long long res = first(a, k / 2);
  res = (res * res) % mod;
  if (k % 2) res = (res * a) % mod;
  return res;
}
int main() {
  cin >> n >> x;
  for (int i = 0; i < n; i++) cin >> a[i];
  long long sum = 0;
  for (int i = 0; i < n; i++) sum += a[i];
  long long ans = first(x, sum - a[n - 1]);
  long long m = a[n - 1];
  for (int i = 0; i < n; i++) a[i] = m - a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    st[a[i]]++;
    second.insert(1ll * a[i]);
  }
  for (__typeof(second.begin()) it = second.begin(); it != second.end(); it++) {
    long long y = *it;
    if (st[y] >= x) {
      st[y + 1] += st[y] / x;
      st[y] %= x;
      second.insert(y + 1);
    }
  }
  long long k;
  for (__typeof(second.begin()) it = second.begin(); it != second.end(); it++) {
    if (st[*it]) {
      k = min(m, *it);
      break;
    }
  }
  if (*second.begin() == 0 && st[0]) {
    k = 0;
  }
  ans = (ans * first(x, k)) % mod;
  cout << ans << endl;
  return 0;
}
