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 double eps = 1.0e-9;
const unsigned long long MAX_N = 100001;
const unsigned long long MAX_M = 1000001;
const long long mod_val = 1000000007;
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long pow_mod(long long base, long long val) {
  if (val == 1) {
    return base % mod_val;
  } else if (val == 0) {
    return 1;
  }
  long long num = 1;
  if (val % 2 == 1) num = base;
  long long t = pow_mod(base, val / 2);
  t = (t * t) % mod_val;
  num = (num * t) % mod_val;
  return num;
}
long long vals[MAX_N];
int main(int argc, char** argv) {
  std::ios::sync_with_stdio(false);
  long long n, x, val, sum = 0;
  cin >> n >> x;
  map<long long, long long> bits;
  for (unsigned int i = 0; i < (unsigned long long)(n); ++i) {
    cin >> vals[i];
    sum += vals[i];
  }
  for (unsigned int i = 0; i < (unsigned long long)(n); ++i) {
    bits[sum - vals[i]]++;
  }
  map<long long, long long>::iterator iter;
  iter = bits.begin();
  while (iter != bits.end()) {
    val = iter->first;
    if (iter->second >= x) {
      bits[val + 1] += iter->second / x;
    }
    iter->second = iter->second % x;
    iter = bits.lower_bound(val + 1);
  }
  bool isFind = false;
  long long min_bit;
  for (iter = bits.begin(); iter != bits.end(); iter++) {
    if (iter->second != 0) {
      if (!isFind) {
        min_bit = iter->first;
        break;
      }
    }
  }
  val = pow_mod(x, min(min_bit, sum)) % mod_val;
  cout << val << endl;
  return 0;
}
