#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
int n, k, a[1000010], l[1000010], r[1000010];
stack<int> s;
long long cal(int x, int y) {
  if (x < 0 || y - x + 1 < k) return 0;
  long long t = (y - x + 2 - k) / (k - 1) + 1;
  long long res = (t * (y - x + 2 - k) % MOD -
                   t * (t - 1) / 2 % MOD * (k - 1) % MOD + MOD) %
                  MOD;
  return res;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    l[i] = i;
    while (!s.empty() && a[s.top()] < a[i]) {
      l[i] = l[s.top()];
      s.pop();
    }
    s.push(i);
  }
  while (!s.empty()) s.pop();
  for (int i = n - 1; i >= 0; i--) {
    r[i] = i;
    while (!s.empty() && a[s.top()] <= a[i]) {
      r[i] = r[s.top()];
      s.pop();
    }
    s.push(i);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long temp = cal(l[i], r[i]) - cal(l[i], i - 1) - cal(i + 1, r[i]);
    temp = temp % MOD * a[i] % MOD;
    ans = (ans + temp) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Please formulate 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;
const int MOD = 1000000007;
int n, k, a[1000010], l[1000010], r[1000010];
stack<int> s;
long long cal(int x, int y) {
  if (x < 0 || y - x + 1 < k) return 0;
  long long t = (y - x + 2 - k) / (k - 1) + 1;
  long long res = (t * (y - x + 2 - k) % MOD -
                   t * (t - 1) / 2 % MOD * (k - 1) % MOD + MOD) %
                  MOD;
  return res;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < n; i++) {
    l[i] = i;
    while (!s.empty() && a[s.top()] < a[i]) {
      l[i] = l[s.top()];
      s.pop();
    }
    s.push(i);
  }
  while (!s.empty()) s.pop();
  for (int i = n - 1; i >= 0; i--) {
    r[i] = i;
    while (!s.empty() && a[s.top()] <= a[i]) {
      r[i] = r[s.top()];
      s.pop();
    }
    s.push(i);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long temp = cal(l[i], r[i]) - cal(l[i], i - 1) - cal(i + 1, r[i]);
    temp = temp % MOD * a[i] % MOD;
    ans = (ans + temp) % MOD;
  }
  printf("%lld\n", ans);
  return 0;
}

```