#include <bits/stdc++.h>
using namespace std;
const int mo = 1e9 + 7;
const int maxn = 1000010;
int n, k, l[maxn], r[maxn], lc;
long long ans, a[maxn], h[maxn];
int main() {
  scanf("%d%d", &n, &k);
  l[1] = 1;
  r[n] = n;
  for (int i = 1; i <= n; i++)
    scanf("%lld", &a[i]), h[i] = (h[i - 1] + 1ll * i / (k - 1)) % mo;
  for (int i = 2; i <= n; i++) {
    lc = i;
    while (lc > 1 && a[lc - 1] <= a[i]) {
      lc = l[lc - 1];
    }
    l[i] = lc;
  }
  for (int i = n - 1; i >= 1; i--) {
    lc = i;
    while (lc < n && a[lc + 1] < a[i]) {
      lc = r[lc + 1];
    }
    r[i] = lc;
  }
  for (int x = 1; x <= n; x++)
    ans = (ans +
           a[x] *
               (h[r[x] - l[x]] - h[r[x] - x - 1] - h[x - l[x] - 1] + 2ll * mo) %
               mo) %
          mo;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int mo = 1e9 + 7;
const int maxn = 1000010;
int n, k, l[maxn], r[maxn], lc;
long long ans, a[maxn], h[maxn];
int main() {
  scanf("%d%d", &n, &k);
  l[1] = 1;
  r[n] = n;
  for (int i = 1; i <= n; i++)
    scanf("%lld", &a[i]), h[i] = (h[i - 1] + 1ll * i / (k - 1)) % mo;
  for (int i = 2; i <= n; i++) {
    lc = i;
    while (lc > 1 && a[lc - 1] <= a[i]) {
      lc = l[lc - 1];
    }
    l[i] = lc;
  }
  for (int i = n - 1; i >= 1; i--) {
    lc = i;
    while (lc < n && a[lc + 1] < a[i]) {
      lc = r[lc + 1];
    }
    r[i] = lc;
  }
  for (int x = 1; x <= n; x++)
    ans = (ans +
           a[x] *
               (h[r[x] - l[x]] - h[r[x] - x - 1] - h[x - l[x] - 1] + 2ll * mo) %
               mo) %
          mo;
  printf("%lld\n", ans);
  return 0;
}

```