#include <bits/stdc++.h>
using namespace std;
stack<int> s;
int n, k, a[1100000], l[1100000], r[1100000];
long long count(int l, int r, int t) {
  int k = (r - l) / t + 1;
  return ((1ll * l + r) * k / 2) % 1000000007;
}
int calc(int x, int y, int z) {
  int n = x + y + z;
  int a = x % (k - 1), b = n % (k - 1), c = z % (k - 1);
  return (count(b, n, k - 1) - count(a, x, k - 1) - count(c, z, k - 1)) %
         1000000007;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  s.push(1);
  for (int i = 2; i <= n; i++) {
    while (s.size() && a[i] >= a[s.top()]) s.pop();
    if (s.size()) l[i] = s.top();
    s.push(i);
  }
  while (s.size()) s.pop();
  r[n] = n + 1;
  s.push(n);
  for (int i = n - 1; i >= 1; i--) {
    while (s.size() && a[i] > a[s.top()]) s.pop();
    if (s.size())
      r[i] = s.top();
    else
      r[i] = n + 1;
    s.push(i);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int x = max(i - l[i] - 1 - k + 1, 0), z = max(r[i] - i - 1 - k + 1, 0);
    int y = max(r[i] - l[i] - 1 - k + 1 - x - z, 0);
    ans = (ans + 1ll * calc(x, y, z) * a[i] % 1000000007) % 1000000007;
  }
  if (ans < 0) ans += 1000000007;
  cout << ans;
  return 0;
}

### Prompt

Generate a Cpp solution 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;
stack<int> s;
int n, k, a[1100000], l[1100000], r[1100000];
long long count(int l, int r, int t) {
  int k = (r - l) / t + 1;
  return ((1ll * l + r) * k / 2) % 1000000007;
}
int calc(int x, int y, int z) {
  int n = x + y + z;
  int a = x % (k - 1), b = n % (k - 1), c = z % (k - 1);
  return (count(b, n, k - 1) - count(a, x, k - 1) - count(c, z, k - 1)) %
         1000000007;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  s.push(1);
  for (int i = 2; i <= n; i++) {
    while (s.size() && a[i] >= a[s.top()]) s.pop();
    if (s.size()) l[i] = s.top();
    s.push(i);
  }
  while (s.size()) s.pop();
  r[n] = n + 1;
  s.push(n);
  for (int i = n - 1; i >= 1; i--) {
    while (s.size() && a[i] > a[s.top()]) s.pop();
    if (s.size())
      r[i] = s.top();
    else
      r[i] = n + 1;
    s.push(i);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int x = max(i - l[i] - 1 - k + 1, 0), z = max(r[i] - i - 1 - k + 1, 0);
    int y = max(r[i] - l[i] - 1 - k + 1 - x - z, 0);
    ans = (ans + 1ll * calc(x, y, z) * a[i] % 1000000007) % 1000000007;
  }
  if (ans < 0) ans += 1000000007;
  cout << ans;
  return 0;
}

```