#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 1e6 + 5;
int n, k, top;
int a[N], l[N], r[N], s[N];
long long ans;
long long calc(int l, int r) {
  if (l > r || r - l + 1 < k) return 0;
  int m = (r - l + 2 - k) / (k - 1);
  long long ret = 1LL * (m + 1) * (r - l + 2 - k) % mod -
                  1LL * (1 + m) * m / 2 % mod * (k - 1) % mod;
  return ret;
}
int main() {
  int i;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (i = 1; i <= n; ++i) {
    while (top && a[s[top]] < a[i]) --top;
    l[i] = s[top] + 1, s[++top] = i;
  }
  top = 0, s[top] = n + 1;
  for (i = n; i; --i) {
    while (top && a[s[top]] <= a[i]) --top;
    r[i] = s[top] - 1, s[++top] = i;
  }
  for (i = 1; i <= n; ++i)
    ans = (ans +
           (calc(l[i], r[i]) - calc(l[i], i - 1) - calc(i + 1, r[i])) * a[i]) %
          mod;
  if (ans < 0) ans += mod;
  printf("%lld\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>
using namespace std;
const long long mod = 1e9 + 7;
const int N = 1e6 + 5;
int n, k, top;
int a[N], l[N], r[N], s[N];
long long ans;
long long calc(int l, int r) {
  if (l > r || r - l + 1 < k) return 0;
  int m = (r - l + 2 - k) / (k - 1);
  long long ret = 1LL * (m + 1) * (r - l + 2 - k) % mod -
                  1LL * (1 + m) * m / 2 % mod * (k - 1) % mod;
  return ret;
}
int main() {
  int i;
  scanf("%d%d", &n, &k);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (i = 1; i <= n; ++i) {
    while (top && a[s[top]] < a[i]) --top;
    l[i] = s[top] + 1, s[++top] = i;
  }
  top = 0, s[top] = n + 1;
  for (i = n; i; --i) {
    while (top && a[s[top]] <= a[i]) --top;
    r[i] = s[top] - 1, s[++top] = i;
  }
  for (i = 1; i <= n; ++i)
    ans = (ans +
           (calc(l[i], r[i]) - calc(l[i], i - 1) - calc(i + 1, r[i])) * a[i]) %
          mod;
  if (ans < 0) ans += mod;
  printf("%lld\n", ans);
  return 0;
}

```