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 mod = 1000000007;
int n, x;
long long tot;
int a[111111];
long long b[111111];
int sum[222222];
long long modExp(long long x, long long m) {
  long long ret = 1;
  while (m) {
    if (m & 1) ret = (x * ret) % mod;
    x = (x * x) % mod;
    m /= 2;
  }
  return ret;
}
int main() {
  while (cin >> n >> x) {
    memset(sum, 0, sizeof sum);
    long long tot = 0;
    int i;
    for (i = 1; i <= n; i++) cin >> a[i];
    for (i = 1; i <= n; i++) tot += a[i];
    for (i = 1; i <= n; i++) b[i] = tot - a[i];
    long long DEC = b[1];
    for (i = 1; i <= n; i++) DEC = min(DEC, b[i]);
    long long res = DEC;
    for (i = 1; i <= n; i++) b[i] -= DEC;
    for (i = 1; i <= n; i++)
      if (b[i] <= 2 * n) sum[b[i]]++;
    for (i = 0; i <= n; i++)
      while (sum[i] >= x) sum[i] -= x, sum[i + 1]++;
    for (i = 0; i <= n + 1; i++)
      if (sum[i]) break;
    res += i;
    res = min(res, tot);
    cout << modExp(x, res) << endl;
  }
  return 0;
}
