#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k, i;
  cin >> n >> k;
  int b[n], c[n], e = 0, f = 0;
  long long r = 0, mod = (1e9) + 7, a[n], d[n + 1];
  stack<int> s;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) {
    while (!s.empty() && a[s.top()] < a[i]) {
      b[s.top()] = i;
      s.pop();
    }
    s.push(i);
  }
  while (!s.empty()) {
    b[s.top()] = n;
    s.pop();
  }
  for (i = n - 1; i >= 0; i--) {
    while (!s.empty() && a[s.top()] <= a[i]) {
      c[s.top()] = i;
      s.pop();
    }
    s.push(i);
  }
  while (!s.empty()) {
    c[s.top()] = -1;
    s.pop();
  }
  for (i = 0; i < k; i++) d[i] = 0;
  k--;
  for (; i <= n; i++) {
    if (e % k == 0) {
      e = 0;
      f++;
    }
    e++;
    d[i] = (d[i - 1] + f) % mod;
  }
  k++;
  for (i = 0; i < n; i++) {
    e = b[i] - c[i] - 1;
    r = (r + (d[e] * a[i]) % mod) % mod;
    e = b[i] - i - 1;
    r = (r + mod - ((d[e] * a[i]) % mod)) % mod;
    e = i - c[i] - 1;
    r = (r + mod - ((d[e] * a[i]) % mod)) % mod;
  }
  cout << r;
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

Given an array a of n integers and an integer k (2 ≤ k ≤ n), where each element of the array is denoted by a_i (0 ≤ i < n). Perform the operation z given below on a and print the value of z(a,k) modulo 10^{9}+7.
    
    
      
    function z(array a, integer k):  
        if length(a) < k:  
            return 0  
        else:  
            b = empty array  
            ans = 0  
            for i = 0 .. (length(a) - k):  
                temp = a[i]  
                for j = i .. (i + k - 1):  
                    temp = max(temp, a[j])  
                append temp to the end of b  
                ans = ans + temp  
            return ans + z(b, k)  
    

Input

The first line of input contains two integers n and k (2 ≤ k ≤ n ≤ 10^6) — the length of the initial array a and the parameter k.

The second line of input contains n integers a_0, a_1, …, a_{n - 1} (1 ≤ a_{i} ≤ 10^9) — the elements of the array a.

Output

Output the only integer, the value of z(a,k) modulo 10^9+7.

Examples

Input

3 2
9 1 10


Output

29


Input

5 3
5 8 7 1 9


Output

34

Note

In the first example: 

  * for a=(9,1,10), ans=19 and b=(9,10), 
  * for a=(9,10), ans=10 and b=(10), 
  * for a=(10), ans=0. 



So the returned value is 19+10+0=29.

In the second example: 

  * for a=(5,8,7,1,9), ans=25 and b=(8,8,9), 
  * for a=(8,8,9), ans=9 and b=(9), 
  * for a=(9), ans=0. 



So the returned value is 25+9+0=34.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k, i;
  cin >> n >> k;
  int b[n], c[n], e = 0, f = 0;
  long long r = 0, mod = (1e9) + 7, a[n], d[n + 1];
  stack<int> s;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++) {
    while (!s.empty() && a[s.top()] < a[i]) {
      b[s.top()] = i;
      s.pop();
    }
    s.push(i);
  }
  while (!s.empty()) {
    b[s.top()] = n;
    s.pop();
  }
  for (i = n - 1; i >= 0; i--) {
    while (!s.empty() && a[s.top()] <= a[i]) {
      c[s.top()] = i;
      s.pop();
    }
    s.push(i);
  }
  while (!s.empty()) {
    c[s.top()] = -1;
    s.pop();
  }
  for (i = 0; i < k; i++) d[i] = 0;
  k--;
  for (; i <= n; i++) {
    if (e % k == 0) {
      e = 0;
      f++;
    }
    e++;
    d[i] = (d[i - 1] + f) % mod;
  }
  k++;
  for (i = 0; i < n; i++) {
    e = b[i] - c[i] - 1;
    r = (r + (d[e] * a[i]) % mod) % mod;
    e = b[i] - i - 1;
    r = (r + mod - ((d[e] * a[i]) % mod)) % mod;
    e = i - c[i] - 1;
    r = (r + mod - ((d[e] * a[i]) % mod)) % mod;
  }
  cout << r;
  return 0;
}

```