#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10, mod = 1e9 + 7;
void add(int& x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
}
int n, k, a[N], q[N], top, l[N], r[N];
int f(int len) {
  int div_begin = max(len - k, 0);
  int div_end = div_begin - div_begin / k * k;
  return ((long long)(div_begin + div_end) * (div_begin / k + 1) >> 1) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  q[top = 0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (; top && a[i] >= a[q[top]]; --top)
      ;
    l[i] = q[top];
    q[++top] = i;
  }
  q[top = 0] = n + 1;
  for (int i = n; i; --i) {
    for (; top && a[i] > a[q[top]]; --top)
      ;
    r[i] = q[top];
    q[++top] = i;
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    add(ans,
        (((long long)f(r[i] - l[i] - 1) - f(i - l[i] - 1) - f(r[i] - i - 1)) %
             mod +
         mod) *
            a[i] % mod);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 N = 1e6 + 10, mod = 1e9 + 7;
void add(int& x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
}
int n, k, a[N], q[N], top, l[N], r[N];
int f(int len) {
  int div_begin = max(len - k, 0);
  int div_end = div_begin - div_begin / k * k;
  return ((long long)(div_begin + div_end) * (div_begin / k + 1) >> 1) % mod;
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  q[top = 0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (; top && a[i] >= a[q[top]]; --top)
      ;
    l[i] = q[top];
    q[++top] = i;
  }
  q[top = 0] = n + 1;
  for (int i = n; i; --i) {
    for (; top && a[i] > a[q[top]]; --top)
      ;
    r[i] = q[top];
    q[++top] = i;
  }
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    add(ans,
        (((long long)f(r[i] - l[i] - 1) - f(i - l[i] - 1) - f(r[i] - i - 1)) %
             mod +
         mod) *
            a[i] % mod);
  }
  printf("%d\n", ans);
  return 0;
}

```