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 long long int mod = 1e9 + 7;
long long int t, n, m, a[1000005] = {0};
map<long long int, long long int> mp;
long long int binexpo(long long int a, long long int b, long long int m) {
  a %= m;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, x, y;
  cin >> n >> x;
  long long int ss = 0;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    ss += a[i];
  }
  for (i = 1; i <= n; i++) mp[ss - a[i]]++;
  while (1) {
    auto it = mp.begin();
    y = it->first;
    long long int gg = (it->second) % x;
    if (gg) break;
    mp[1 + y] += ((it->second) / x);
    mp[y] -= ((it->second) / x) * x;
    if (mp[y] == 0) mp.erase(y);
  }
  auto it = mp.begin();
  y = min(ss, it->first);
  y = binexpo(x, y, mod);
  cout << y;
  return 0;
}
