#include <bits/stdc++.h>
using namespace std;
int N, K;
long long ans = 0;
int arr[1000005];
pair<int, int> srt[1000005];
set<int> st;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> K;
  for (int i = 1; i <= N; i++) {
    cin >> arr[i];
    ans -= arr[i];
    srt[i] = {arr[i], i};
  }
  sort(srt + 1, srt + 1 + N, greater<pair<int, int>>());
  st.insert(0), st.insert(N + 1);
  for (int i = 1; i <= N; i++) {
    auto it = st.lower_bound(srt[i].second);
    int r = *it, l = *(--it);
    r--, l++;
    int p = srt[i].second, v = srt[i].first;
    int mn = min(r - p, p - l), mx = max(r - p, p - l);
    mn++, mx++;
    long long l1 = (mn - 1) / (K - 1);
    long long cnt = (2 + l1 * (K - 1)) * (l1 + 1) / 2;
    long long l2 = (mx - 1) / (K - 1);
    cnt += (l2 - l1) * (mn);
    long long l3 = (r - l) / (K - 1);
    if (l2 != l3) {
      long long k = (r - l + 1) - (l3) * (K - 1);
      cnt += (2 * k + (l3 - l2 - 1) * (K - 1)) * (l3 - l2) / 2;
    }
    ans += cnt % 1000000007 * v;
    ans %= 1000000007;
    st.insert(p);
  }
  cout << ans << "\n";
}

### Prompt

Please provide a cpp coded solution 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;
long long ans = 0;
int arr[1000005];
pair<int, int> srt[1000005];
set<int> st;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> K;
  for (int i = 1; i <= N; i++) {
    cin >> arr[i];
    ans -= arr[i];
    srt[i] = {arr[i], i};
  }
  sort(srt + 1, srt + 1 + N, greater<pair<int, int>>());
  st.insert(0), st.insert(N + 1);
  for (int i = 1; i <= N; i++) {
    auto it = st.lower_bound(srt[i].second);
    int r = *it, l = *(--it);
    r--, l++;
    int p = srt[i].second, v = srt[i].first;
    int mn = min(r - p, p - l), mx = max(r - p, p - l);
    mn++, mx++;
    long long l1 = (mn - 1) / (K - 1);
    long long cnt = (2 + l1 * (K - 1)) * (l1 + 1) / 2;
    long long l2 = (mx - 1) / (K - 1);
    cnt += (l2 - l1) * (mn);
    long long l3 = (r - l) / (K - 1);
    if (l2 != l3) {
      long long k = (r - l + 1) - (l3) * (K - 1);
      cnt += (2 * k + (l3 - l2 - 1) * (K - 1)) * (l3 - l2) / 2;
    }
    ans += cnt % 1000000007 * v;
    ans %= 1000000007;
    st.insert(p);
  }
  cout << ans << "\n";
}

```