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 p = 1000000007ll, x;
long long binpow(long long v) {
  if (v == 0ll) return 1ll;
  if (v == 1ll) return x;
  if (v % 2ll == 0ll)
    return ((binpow(v / 2ll) % p) * (binpow(v / 2ll) % p)) % p;
  return (binpow(v - 1ll) * x) % p;
}
long long s = 0ll, a[111111], y[111111], S = 0ll, Y, z = 0;
int n;
int main() {
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    s += a[i];
  }
  for (int i = 0; i < n; i++) y[i] = s - a[i];
  Y = y[n - 1];
  for (int i = 0; i < n; i++) y[i] -= Y;
  if (a[n - 1] == 0 || n == 1) {
    cout << "1";
    return 0;
  }
  Y = binpow(Y);
  for (int i = 0; i < n; i++)
    if (!y[i]) S++;
  if (S % x) {
    cout << Y;
    return 0;
  }
  Y = (Y * x) % p;
  a[n - 1]--;
  if (!a[n - 1]) {
    cout << Y;
    return 0;
  }
  for (int i = 0; i < n; i++)
    if (!(y[i] - 1)) {
      z++;
    }
  int w = 2;
  while (S % x == 0 && a[n - 1]) {
    S /= x;
    S += z;
    z = 0;
    for (int i = 0; i < n; i++)
      if (y[i] - w == 0) z++;
    if (S % x == 0) w++, Y = (Y * x) % p, a[n - 1]--;
  }
  cout << Y % p;
  return 0;
}
