#include <bits/stdc++.h>
using namespace std;
int n, k;
int a[1000309];
int nxt[1000309], prv[1000309], dp[1000309];
int main() {
  while (scanf("%d %d", &n, &k) != EOF) {
    dp[0] = 0;
    for (int i = 1; (i) <= int(n); (i)++) {
      dp[i] = (dp[i - 1] + (i - 1) / (k - 1)) % 1000000007;
    }
    for (int i = 0; (i) < int(n); (i)++) scanf("%d", &a[i]);
    stack<int> sp, sn;
    for (int i = 0; (i) < int(n); (i)++) {
      while (!sp.empty() && a[sp.top()] < a[i]) sp.pop();
      prv[i] = sp.empty() ? -1 : sp.top();
      sp.push(i);
    }
    for (int i = int(n) - 1; (i) >= 0; (i)--) {
      while (!sn.empty() && a[sn.top()] <= a[i]) sn.pop();
      nxt[i] = sn.empty() ? n : sn.top();
      sn.push(i);
    }
    int ans = 0;
    for (int i = 0; (i) < int(n); (i)++) {
      int r = nxt[i];
      int l = prv[i];
      int cur = ((dp[r - l - 1] - dp[i - l - 1] - dp[r - i - 1]) % 1000000007 +
                 1000000007) %
                1000000007;
      cur = (a[i] * 1ll * cur) % 1000000007;
      ans = (ans + cur) % 1000000007;
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
int n, k;
int a[1000309];
int nxt[1000309], prv[1000309], dp[1000309];
int main() {
  while (scanf("%d %d", &n, &k) != EOF) {
    dp[0] = 0;
    for (int i = 1; (i) <= int(n); (i)++) {
      dp[i] = (dp[i - 1] + (i - 1) / (k - 1)) % 1000000007;
    }
    for (int i = 0; (i) < int(n); (i)++) scanf("%d", &a[i]);
    stack<int> sp, sn;
    for (int i = 0; (i) < int(n); (i)++) {
      while (!sp.empty() && a[sp.top()] < a[i]) sp.pop();
      prv[i] = sp.empty() ? -1 : sp.top();
      sp.push(i);
    }
    for (int i = int(n) - 1; (i) >= 0; (i)--) {
      while (!sn.empty() && a[sn.top()] <= a[i]) sn.pop();
      nxt[i] = sn.empty() ? n : sn.top();
      sn.push(i);
    }
    int ans = 0;
    for (int i = 0; (i) < int(n); (i)++) {
      int r = nxt[i];
      int l = prv[i];
      int cur = ((dp[r - l - 1] - dp[i - l - 1] - dp[r - i - 1]) % 1000000007 +
                 1000000007) %
                1000000007;
      cur = (a[i] * 1ll * cur) % 1000000007;
      ans = (ans + cur) % 1000000007;
    }
    printf("%d\n", ans);
  }
  return 0;
}

```