#include <bits/stdc++.h>
using namespace std;
const int mo = 1000000007;
const int N = 1000005;
int k, n, a[N], l[N], r[N], q[N], ans;
int F(int l) {
  int pr = 5000000 / k * k;
  int pl = l + (k - l % k) % k;
  return 1ll * (pr + pl - 2 * l + 2) * (pr / k - pl / k + 1) / 2 % mo;
}
int Func(int vl, int vr) {
  int ans = F(0);
  ans = (ans + mo - F(vl)) % mo;
  ans = (ans + mo - F(vr)) % mo;
  ans = (ans + F(vl + vr)) % mo;
  return ans + mo - 1;
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i]);
  for (int i = (int)(1); i <= (int)(n); i++) l[i] = 0, r[i] = n + 1;
  int t = 1;
  q[1] = 1;
  for (int i = (int)(2); i <= (int)(n); i++) {
    for (; t && a[q[t]] <= a[i]; --t) r[q[t]] = i;
    l[i] = q[t];
    q[++t] = i;
  }
  for (int i = (int)(1); i <= (int)(n); i++) {
    ans = (ans + 1ll * a[i] * Func(i - l[i], r[i] - i)) % mo;
  }
  cout << ans << 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 mo = 1000000007;
const int N = 1000005;
int k, n, a[N], l[N], r[N], q[N], ans;
int F(int l) {
  int pr = 5000000 / k * k;
  int pl = l + (k - l % k) % k;
  return 1ll * (pr + pl - 2 * l + 2) * (pr / k - pl / k + 1) / 2 % mo;
}
int Func(int vl, int vr) {
  int ans = F(0);
  ans = (ans + mo - F(vl)) % mo;
  ans = (ans + mo - F(vr)) % mo;
  ans = (ans + F(vl + vr)) % mo;
  return ans + mo - 1;
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (int i = (int)(1); i <= (int)(n); i++) scanf("%d", &a[i]);
  for (int i = (int)(1); i <= (int)(n); i++) l[i] = 0, r[i] = n + 1;
  int t = 1;
  q[1] = 1;
  for (int i = (int)(2); i <= (int)(n); i++) {
    for (; t && a[q[t]] <= a[i]; --t) r[q[t]] = i;
    l[i] = q[t];
    q[++t] = i;
  }
  for (int i = (int)(1); i <= (int)(n); i++) {
    ans = (ans + 1ll * a[i] * Func(i - l[i], r[i] - i)) % mo;
  }
  cout << ans << endl;
}

```