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 mod(long b, long long p, long long m) {
  if (p == 0) {
    return 1;
  } else if (p % 2 == 0) {
    long long x = mod(b, p / 2, m);
    return (x * x) % m;
  } else {
    return ((b % m) * mod(b, p - 1, m)) % m;
  }
}
int main(void) {
  ios_base::sync_with_stdio(0);
  long long n, x;
  cin >> n >> x;
  vector<long long> v(n);
  long long mx = -1, sum = 0;
  for (int i = 0; i < (n); i++) {
    cin >> v[i];
    sum += v[i];
    mx = max(mx, v[i]);
  }
  sum -= mx;
  for (int i = 0; i < (n); i++) {
    v[i] = mx - v[i];
  }
  sort((v).begin(), (v).end());
  reverse((v).begin(), (v).end());
  vector<long long> re;
  stack<pair<long long, long long> > st;
  for (int i = 0; i < (((int)(v).size())); i++) {
    int cc = 0;
    for (int j = (i); j < (((int)(v).size())); j++) {
      if (v[i] != v[j]) break;
      cc++;
    }
    st.push(make_pair(v[i], cc));
    i += cc - 1;
  }
  while (((int)(st).size())) {
    long long t = st.top().first;
    long long cc = st.top().second;
    st.pop();
    while (((int)(st).size()) && st.top().first == t) {
      cc += st.top().second;
      st.pop();
    }
    if (cc >= x) st.push(make_pair(t + 1, cc / x));
    if (cc % x) {
      re.push_back(t);
      break;
    }
  }
  sort((re).begin(), (re).end());
  cout << mod(x, min(re[0], mx) + sum, 1000000007);
  return 0;
}
