#include <bits/stdc++.h>
const int N = 1e6 + 5, md = 1e9 + 7;
int L[N], R[N], q[N], top, n, k, a[N], ans = 0;
int main() {
  scanf("%d%d", &n, &k), --k;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  L[1] = q[1] = top = 1;
  for (int i = 2; i <= n; ++i) {
    while (top && a[i] > a[q[top]]) --top;
    L[i] = q[top] + 1;
    q[++top] = i;
  }
  R[n] = q[top = 1] = n, *q = n + 1;
  for (int i = n - 1; i; --i) {
    while (top && a[i] >= a[q[top]]) --top;
    R[i] = q[top] - 1;
    q[++top] = i;
  }
  for (int d = 1; d <= n; ++d) {
    const int &l = L[d], &r = R[d];
    if ((l + r >> 1) > d)
      for (int i = l; i <= d; ++i)
        ans = (ans + ((r - i) / k - (d - i - (d > i)) / k) * (long long)a[d]) %
              md;
    else
      for (int i = r; i >= d; --i)
        ans = (ans + ((i - l) / k - (i - d - (i > d)) / k) * (long long)a[d]) %
              md;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 N = 1e6 + 5, md = 1e9 + 7;
int L[N], R[N], q[N], top, n, k, a[N], ans = 0;
int main() {
  scanf("%d%d", &n, &k), --k;
  for (int i = 1; i <= n; ++i) scanf("%d", a + i);
  L[1] = q[1] = top = 1;
  for (int i = 2; i <= n; ++i) {
    while (top && a[i] > a[q[top]]) --top;
    L[i] = q[top] + 1;
    q[++top] = i;
  }
  R[n] = q[top = 1] = n, *q = n + 1;
  for (int i = n - 1; i; --i) {
    while (top && a[i] >= a[q[top]]) --top;
    R[i] = q[top] - 1;
    q[++top] = i;
  }
  for (int d = 1; d <= n; ++d) {
    const int &l = L[d], &r = R[d];
    if ((l + r >> 1) > d)
      for (int i = l; i <= d; ++i)
        ans = (ans + ((r - i) / k - (d - i - (d > i)) / k) * (long long)a[d]) %
              md;
    else
      for (int i = r; i >= d; --i)
        ans = (ans + ((i - l) / k - (i - d - (i > d)) / k) * (long long)a[d]) %
              md;
  }
  printf("%d\n", ans);
  return 0;
}

```