#include <bits/stdc++.h>
using namespace std;
const int N = 1E6 + 5;
const int MOD = 1E9 + 7;
int n, k, f[N], pre[N], ne[N], b[N], a[N], x;
long long ans;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n + 1; i++) f[i] = (f[i - 1] + (i - 1) / (k - 1)) % MOD;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = i;
    pre[i] = i - 1;
    ne[i] = i + 1;
  }
  sort(b + 1, b + n + 1, [](int x, int y) { return a[x] < a[y]; });
  for (int i = 1; i <= n; i++) {
    x = b[i];
    (ans += (long long)a[x] * (f[ne[x] - pre[x] - 1] - f[ne[x] - x - 1] -
                               f[x - pre[x] - 1])) %= MOD;
    pre[ne[x]] = pre[x];
    ne[pre[x]] = ne[x];
  }
  cout << (ans + MOD) % MOD << endl;
}

### 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 N = 1E6 + 5;
const int MOD = 1E9 + 7;
int n, k, f[N], pre[N], ne[N], b[N], a[N], x;
long long ans;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n + 1; i++) f[i] = (f[i - 1] + (i - 1) / (k - 1)) % MOD;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = i;
    pre[i] = i - 1;
    ne[i] = i + 1;
  }
  sort(b + 1, b + n + 1, [](int x, int y) { return a[x] < a[y]; });
  for (int i = 1; i <= n; i++) {
    x = b[i];
    (ans += (long long)a[x] * (f[ne[x] - pre[x] - 1] - f[ne[x] - x - 1] -
                               f[x - pre[x] - 1])) %= MOD;
    pre[ne[x]] = pre[x];
    ne[pre[x]] = ne[x];
  }
  cout << (ans + MOD) % MOD << endl;
}

```