#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000100;
const long long MOD = 1e9 + 7;
int N, K;
int arr[MAXN];
int nprev[MAXN], nnext[MAXN];
pair<int, int> ord[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> K;
  for (int i = 0; i < N; i++) {
    cin >> arr[i];
    ord[i] = make_pair(arr[i], i);
  }
  sort(ord, ord + N);
  for (int i = 0; i < N; i++) {
    nprev[i] = i - 1;
    nnext[i] = i + 1;
  }
  for (int i = 0; i < N; i++) {
    int cloc = ord[i].second;
    if (nprev[cloc] >= 0) nnext[nprev[cloc]] = nnext[cloc];
    if (nnext[cloc] < N) nprev[nnext[cloc]] = nprev[cloc];
  }
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    long long l = i - nprev[i], r = nnext[i] - i;
    if (l > r) swap(l, r);
    long long c1 = (l + K - 3) / (K - 1);
    long long c2 = (r + K - 3) / (K - 1);
    long long c3 = (l + r + K - 3) / (K - 1);
    long long ntot = 0;
    ntot += c1 + ((K - 1) * (c1) * (c1 - 1)) / 2;
    ntot += (c2 - c1) * l;
    ntot += (c3 - c2) * (l + r);
    ntot += c2 + ((K - 1) * (c2) * (c2 - 1)) / 2;
    ;
    ntot -= c3 + ((K - 1) * (c3) * (c3 - 1)) / 2;
    ntot %= MOD;
    ans = (ans + ntot * arr[i]) % MOD;
  }
  for (int i = 0; i < N; i++) ans = (ans - arr[i]) % MOD;
  ans = (ans + MOD) % MOD;
  cout << ans << "\n";
}

### 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 int MAXN = 1000100;
const long long MOD = 1e9 + 7;
int N, K;
int arr[MAXN];
int nprev[MAXN], nnext[MAXN];
pair<int, int> ord[MAXN];
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> K;
  for (int i = 0; i < N; i++) {
    cin >> arr[i];
    ord[i] = make_pair(arr[i], i);
  }
  sort(ord, ord + N);
  for (int i = 0; i < N; i++) {
    nprev[i] = i - 1;
    nnext[i] = i + 1;
  }
  for (int i = 0; i < N; i++) {
    int cloc = ord[i].second;
    if (nprev[cloc] >= 0) nnext[nprev[cloc]] = nnext[cloc];
    if (nnext[cloc] < N) nprev[nnext[cloc]] = nprev[cloc];
  }
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    long long l = i - nprev[i], r = nnext[i] - i;
    if (l > r) swap(l, r);
    long long c1 = (l + K - 3) / (K - 1);
    long long c2 = (r + K - 3) / (K - 1);
    long long c3 = (l + r + K - 3) / (K - 1);
    long long ntot = 0;
    ntot += c1 + ((K - 1) * (c1) * (c1 - 1)) / 2;
    ntot += (c2 - c1) * l;
    ntot += (c3 - c2) * (l + r);
    ntot += c2 + ((K - 1) * (c2) * (c2 - 1)) / 2;
    ;
    ntot -= c3 + ((K - 1) * (c3) * (c3 - 1)) / 2;
    ntot %= MOD;
    ans = (ans + ntot * arr[i]) % MOD;
  }
  for (int i = 0; i < N; i++) ans = (ans - arr[i]) % MOD;
  ans = (ans + MOD) % MOD;
  cout << ans << "\n";
}

```