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 = 1000000111;
const int MOD = 1000000007;
const int S_N = 100010;
template <typename A, typename B>
ostream& operator<<(ostream& out, pair<A, B>& p) {
  cout << "(" << p.first << "," << p.second << ") ";
}
template <typename A, typename B>
void printP(const pair<A, B>& p, const char* str = NULL) {
  if (str != NULL) cout << str << ":";
  cout << "(" << p.first << "," << p.second << ") ";
}
template <typename T>
void printV(const vector<T>& v, const char* str = NULL) {
  if (str != NULL) cout << str << ": ";
  for (int i = 0; i < (v.size()); ++i) cout << v[i] << " ";
  cout << endl;
}
long long powd(long long x, long long p) {
  if (p == 0) return 1;
  long long r = powd(x, p / 2);
  r = ((r * r) % MOD);
  if (p % 2) r = ((r * x) % MOD);
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, x;
  cin >> n >> x;
  long long maxi = 0, sum = 0;
  vector<long long> A(n);
  for (int i = 0; i < (n); ++i) {
    cin >> A[i];
    maxi = max(maxi, A[i]);
    sum += A[i];
  }
  vector<int> C(40);
  for (int i = 0; i < (n); ++i) {
    int p = maxi - A[i];
    if (p < 40) C[p]++;
  }
  int p = 0;
  while (C[p] % x == 0) {
    C[p + 1] += C[p] / x;
    p++;
  }
  p = min(p, (int)maxi);
  long long pow = sum - maxi + p;
  long long res = powd(x, pow);
  cout << res << endl;
  return 0;
}
