#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int a[N], l[N], r[N];
stack<int> st;
int main() {
  int n, k, ans, minx, maxx;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  while (!st.empty()) st.pop();
  for (int i = 1; i <= n; i++) {
    while (!st.empty() && a[st.top()] < a[i]) st.pop();
    if (st.empty())
      l[i] = 1;
    else
      l[i] = st.top() + 1;
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = n; i >= 1; i--) {
    while (!st.empty() && a[st.top()] <= a[i]) st.pop();
    if (st.empty())
      r[i] = n;
    else
      r[i] = st.top() - 1;
    st.push(i);
  }
  ans = 0;
  for (int i = 1; i <= n; i++) {
    if (i - l[i] < r[i] - i) {
      for (int j = l[i]; j <= i; j++) {
        minx = max(i - j, k - 1);
        maxx = r[i] - j + 1;
        if (minx < maxx) {
          ans += 1LL *
                 (max(0, maxx - 1) / (k - 1) - max(0, minx - 1) / (k - 1)) *
                 a[i] % mod;
          ans %= mod;
        }
      }
    } else {
      for (int j = i; j <= r[i]; j++) {
        minx = max(j - i, k - 1);
        maxx = j - l[i] + 1;
        if (minx < maxx) {
          ans += 1LL *
                 (max(0, maxx - 1) / (k - 1) - max(0, minx - 1) / (k - 1)) *
                 a[i] % mod;
          ans %= mod;
        }
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Generate 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 + 10;
const int mod = 1e9 + 7;
int a[N], l[N], r[N];
stack<int> st;
int main() {
  int n, k, ans, minx, maxx;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  while (!st.empty()) st.pop();
  for (int i = 1; i <= n; i++) {
    while (!st.empty() && a[st.top()] < a[i]) st.pop();
    if (st.empty())
      l[i] = 1;
    else
      l[i] = st.top() + 1;
    st.push(i);
  }
  while (!st.empty()) st.pop();
  for (int i = n; i >= 1; i--) {
    while (!st.empty() && a[st.top()] <= a[i]) st.pop();
    if (st.empty())
      r[i] = n;
    else
      r[i] = st.top() - 1;
    st.push(i);
  }
  ans = 0;
  for (int i = 1; i <= n; i++) {
    if (i - l[i] < r[i] - i) {
      for (int j = l[i]; j <= i; j++) {
        minx = max(i - j, k - 1);
        maxx = r[i] - j + 1;
        if (minx < maxx) {
          ans += 1LL *
                 (max(0, maxx - 1) / (k - 1) - max(0, minx - 1) / (k - 1)) *
                 a[i] % mod;
          ans %= mod;
        }
      }
    } else {
      for (int j = i; j <= r[i]; j++) {
        minx = max(j - i, k - 1);
        maxx = j - l[i] + 1;
        if (minx < maxx) {
          ans += 1LL *
                 (max(0, maxx - 1) / (k - 1) - max(0, minx - 1) / (k - 1)) *
                 a[i] % mod;
          ans %= mod;
        }
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}

```