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>
template <typename T>
T gcd(T a, T b) {
  if (!b) return a;
  return gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
template <typename T>
void chmin(T& a, T b) {
  a = (a > b) ? b : a;
}
template <typename T>
void chmax(T& a, T b) {
  a = (a < b) ? b : a;
}
int in() {
  int x;
  scanf("%d", &x);
  return x;
}
using namespace std;
const int MAXN = 100005;
const int MOD = 1000000007LL;
int N, X;
long long P[MAXN];
long long modpow(long long a, long long n) {
  long long res(1);
  while (n) {
    if (n & 1) {
      res = (res * a) % MOD;
    }
    a = (a * 1ll * a) % MOD;
    n >>= 1;
  }
  return res % MOD;
}
int main(void) {
  cin >> N >> X;
  long long denom_exp = 0;
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    cin >> P[i];
    denom_exp += P[i];
  }
  vector<long long> pw;
  for (int i = 0; i < N; i++) {
    pw.push_back(denom_exp - P[i]);
  }
  sort(pw.rbegin(), pw.rend());
  while (true) {
    long long smallest_power = pw.back();
    int cnt = 0;
    while (!pw.empty() && pw.back() == smallest_power) {
      cnt += 1;
      pw.pop_back();
    }
    ans = smallest_power;
    if (cnt % X != 0) break;
    cnt /= X;
    for (int i = 0; i < cnt; i++) {
      pw.push_back(smallest_power + 1);
    }
  }
  cout << modpow(X, min(ans, denom_exp)) << "\n";
  return 0;
}
