#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 69;
long long n, k, l[N], a[N], r[N], mod = 1e9 + 7, res;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    l[i] = i;
    while (l[i] > 1 && a[l[i] - 1] <= a[i]) l[i] = l[l[i] - 1];
  }
  for (int i = n; i >= 1; i--) {
    r[i] = i;
    while (r[i] < n && a[r[i] + 1] < a[i]) r[i] = r[r[i] + 1];
  }
  for (int i = 1; i <= n; i++) {
    long long x = i - l[i], y = r[i] - i;
    long long q = (x + y + 1) / (k - 1);
    long long tmp = 0;
    tmp = (tmp + q - q * (q + 1) / 2 * (k - 1) % mod + mod) % mod;
    long long xq = x / (k - 1);
    tmp = (tmp + xq * (xq + 1) / 2 * (k - 1) + (q - xq) * x) % mod;
    long long yq = y / (k - 1);
    tmp = (tmp + yq * (yq + 1) / 2 * (k - 1) + (q - yq) * y) % mod;
    res = (res + tmp * a[i]) % mod;
  }
  cout << res;
}

### 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 N = 1e6 + 69;
long long n, k, l[N], a[N], r[N], mod = 1e9 + 7, res;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    l[i] = i;
    while (l[i] > 1 && a[l[i] - 1] <= a[i]) l[i] = l[l[i] - 1];
  }
  for (int i = n; i >= 1; i--) {
    r[i] = i;
    while (r[i] < n && a[r[i] + 1] < a[i]) r[i] = r[r[i] + 1];
  }
  for (int i = 1; i <= n; i++) {
    long long x = i - l[i], y = r[i] - i;
    long long q = (x + y + 1) / (k - 1);
    long long tmp = 0;
    tmp = (tmp + q - q * (q + 1) / 2 * (k - 1) % mod + mod) % mod;
    long long xq = x / (k - 1);
    tmp = (tmp + xq * (xq + 1) / 2 * (k - 1) + (q - xq) * x) % mod;
    long long yq = y / (k - 1);
    tmp = (tmp + yq * (yq + 1) / 2 * (k - 1) + (q - yq) * y) % mod;
    res = (res + tmp * a[i]) % mod;
  }
  cout << res;
}

```