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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse4")
using namespace std;
long long power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y % 2) res = (res % 1000000007 * x % 1000000007) % 1000000007;
    x = (x % 1000000007 * x % 1000000007) % 1000000007;
    y = y >> 1;
  }
  return res;
}
long long a[100005], pre[100005];
long long n, x;
long long cal(long long t) {
  long long cnt = 0;
  while (t > 0 && (t % x == 0)) {
    t = t / x;
    cnt++;
  }
  return cnt;
}
map<long long, long long> mpp;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> x;
  long long i;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  long long tot = 0;
  for (i = 1; i <= n; i++) tot = (tot + a[i]);
  for (i = 1; i <= n; i++) {
    pre[i] = (tot - a[i]);
    mpp[pre[i]]++;
  }
  long long sum = 0;
  long long cur = tot;
  auto it = mpp.begin();
  while (it != mpp.end()) {
    long long t = it->first;
    long long y = it->second;
    long long cnt = 0;
    while (y > 0 && y % x == 0) {
      y = y / x;
      cnt++;
    }
    if (cnt)
      mpp[t + cnt] += y;
    else {
      cur = min(cur, t);
      break;
    }
    it++;
  }
  sum = (sum % 1000000006 + cur % 1000000006) % 1000000006;
  long long p = power(x, sum);
  cout << p;
  return 0;
}
