#include <bits/stdc++.h>
using namespace std;
using namespace std;
int a[2000000], n, k;
set<int> dic;
int q[2000000];
bool cmp(int x, int y) { return a[x] > a[y]; }
long long get(int x) {
  if (x <= 0) return 0;
  return (1ll * (1ll * x + x - 1ll * (x / k) * k) * ((x / k) + 1) / 2) %
         1000000007;
}
int main() {
  cin >> n >> k;
  k--;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  dic.clear();
  dic.insert(0);
  dic.insert(n + 1);
  for (int i = 1; i <= n; i++) q[i] = i;
  sort(q + 1, q + n + 1, cmp);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    dic.insert(q[i]);
    int pre;
    auto it1 = dic.find(q[i]);
    --it1;
    pre = *it1;
    int suc;
    auto it2 = dic.find(q[i]);
    ++it2;
    suc = *it2;
    int t1 = max(pre + 1, q[i] - k);
    int t2 = min(q[i], suc - k - 1);
    if (t1 <= t2) {
      long long tmp1 = get(suc - k - 1 - pre);
      long long tmp2 = get(suc - k - 1 - q[i]);
      long long tmp3 = get(q[i] - k - 1 - pre);
      tmp1 = tmp1 - tmp2;
      tmp1 = ((tmp1 % 1000000007) + 1000000007) % 1000000007;
      long long tmp = tmp1 - tmp3;
      tmp = ((tmp % 1000000007) + 1000000007) % 1000000007;
      tmp %= 1000000007;
      ans = (ans + 1ll * tmp * a[q[i]]) % 1000000007;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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;
using namespace std;
int a[2000000], n, k;
set<int> dic;
int q[2000000];
bool cmp(int x, int y) { return a[x] > a[y]; }
long long get(int x) {
  if (x <= 0) return 0;
  return (1ll * (1ll * x + x - 1ll * (x / k) * k) * ((x / k) + 1) / 2) %
         1000000007;
}
int main() {
  cin >> n >> k;
  k--;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  dic.clear();
  dic.insert(0);
  dic.insert(n + 1);
  for (int i = 1; i <= n; i++) q[i] = i;
  sort(q + 1, q + n + 1, cmp);
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    dic.insert(q[i]);
    int pre;
    auto it1 = dic.find(q[i]);
    --it1;
    pre = *it1;
    int suc;
    auto it2 = dic.find(q[i]);
    ++it2;
    suc = *it2;
    int t1 = max(pre + 1, q[i] - k);
    int t2 = min(q[i], suc - k - 1);
    if (t1 <= t2) {
      long long tmp1 = get(suc - k - 1 - pre);
      long long tmp2 = get(suc - k - 1 - q[i]);
      long long tmp3 = get(q[i] - k - 1 - pre);
      tmp1 = tmp1 - tmp2;
      tmp1 = ((tmp1 % 1000000007) + 1000000007) % 1000000007;
      long long tmp = tmp1 - tmp3;
      tmp = ((tmp % 1000000007) + 1000000007) % 1000000007;
      tmp %= 1000000007;
      ans = (ans + 1ll * tmp * a[q[i]]) % 1000000007;
    }
  }
  cout << ans << endl;
  return 0;
}

```