#include <bits/stdc++.h>
using namespace std;
int n, k;
int arr[1000000];
int leftbound[1000000];
int rightbound[1000000];
long long alook[1000001];
long long ccontrib(long long c, int len) {
  long long finc = (len - 1) / (k - 1);
  long long ans = (finc - c) / 2 * (len - finc * k - 1 + len - c * k - 1);
  if (c % 2 == 1) {
    ans += (len - (finc + c) / 2);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    leftbound[i] = -1;
    rightbound[i] = n;
  }
  stack<pair<int, int>> s;
  for (int i = 0; i < n; i++) {
    while (!s.empty() && s.top().first < arr[i]) {
      rightbound[s.top().second] = i;
      s.pop();
    }
    s.push({arr[i], i});
  }
  s = stack<pair<int, int>>();
  for (int i = n - 1; i >= 0; i--) {
    while (!s.empty() && s.top().first <= arr[i]) {
      leftbound[s.top().second] = i;
      s.pop();
    }
    s.push({arr[i], i});
  }
  int numk = 0;
  for (int i = 2; i <= n; i++) {
    if ((i - 1) % (k - 1) == 0) {
      numk++;
    }
    alook[i] = (alook[i - 1] + numk) % 1000000007ll;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int l = i - leftbound[i];
    int r = rightbound[i] - i;
    ans += (alook[l + r - 1] - alook[l - 1] - alook[r - 1]) * arr[i];
    ans = ans % 1000000007ll;
  }
  cout << ans << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 n, k;
int arr[1000000];
int leftbound[1000000];
int rightbound[1000000];
long long alook[1000001];
long long ccontrib(long long c, int len) {
  long long finc = (len - 1) / (k - 1);
  long long ans = (finc - c) / 2 * (len - finc * k - 1 + len - c * k - 1);
  if (c % 2 == 1) {
    ans += (len - (finc + c) / 2);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    leftbound[i] = -1;
    rightbound[i] = n;
  }
  stack<pair<int, int>> s;
  for (int i = 0; i < n; i++) {
    while (!s.empty() && s.top().first < arr[i]) {
      rightbound[s.top().second] = i;
      s.pop();
    }
    s.push({arr[i], i});
  }
  s = stack<pair<int, int>>();
  for (int i = n - 1; i >= 0; i--) {
    while (!s.empty() && s.top().first <= arr[i]) {
      leftbound[s.top().second] = i;
      s.pop();
    }
    s.push({arr[i], i});
  }
  int numk = 0;
  for (int i = 2; i <= n; i++) {
    if ((i - 1) % (k - 1) == 0) {
      numk++;
    }
    alook[i] = (alook[i - 1] + numk) % 1000000007ll;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    int l = i - leftbound[i];
    int r = rightbound[i] - i;
    ans += (alook[l + r - 1] - alook[l - 1] - alook[r - 1]) * arr[i];
    ans = ans % 1000000007ll;
  }
  cout << ans << endl;
}

```