#include <bits/stdc++.h>
const int maxn = 1000010, mod = 1000000007;
int n, k, a[maxn], s[maxn], *t = s, l[maxn], r[maxn];
int main() {
  scanf("%d%d", &n, &k);
  --k;
  *t = -1;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    while (t > s && a[*t] < a[i]) r[*t--] = i;
    l[i] = *t;
    *++t = i;
  }
  while (t > s) r[*t--] = n;
  int sum = 0;
  for (int i = 0; i < n; i++) {
    int x = i - l[i] - 1, y = r[i] - i - 1, u, v, w;
    if (x > y) w = x, x = y, y = w;
    u = x / k, v = y / k, w = (x + y) / k;
    sum = (sum + (u * (u + 1ll) / 2 * k + u + (v - u) * (x + 1ll) -
                  (v + w + 1ll) * (w - v) / 2 * k + (x + y + 1ll) * (w - v)) %
                     mod * a[i]) %
          mod;
  }
  printf("%d\n", sum);
  return 0;
}

### 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>
const int maxn = 1000010, mod = 1000000007;
int n, k, a[maxn], s[maxn], *t = s, l[maxn], r[maxn];
int main() {
  scanf("%d%d", &n, &k);
  --k;
  *t = -1;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    while (t > s && a[*t] < a[i]) r[*t--] = i;
    l[i] = *t;
    *++t = i;
  }
  while (t > s) r[*t--] = n;
  int sum = 0;
  for (int i = 0; i < n; i++) {
    int x = i - l[i] - 1, y = r[i] - i - 1, u, v, w;
    if (x > y) w = x, x = y, y = w;
    u = x / k, v = y / k, w = (x + y) / k;
    sum = (sum + (u * (u + 1ll) / 2 * k + u + (v - u) * (x + 1ll) -
                  (v + w + 1ll) * (w - v) / 2 * k + (x + y + 1ll) * (w - v)) %
                     mod * a[i]) %
          mod;
  }
  printf("%d\n", sum);
  return 0;
}

```