#include <bits/stdc++.h>
using std::cin;
using std::cout;
const int maxn = 1000003, INF = 0x3f3f3f3f, mod = 1e9 + 7;
typedef int arr[maxn];
arr a, l, r, stk;
int main() {
  std::ios::sync_with_stdio(0);
  cin.tie(0);
  int n, K;
  cin >> n >> K;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  int top = 0;
  for (int i = (1); i <= (n); ++i) {
    while (top > 0 && a[stk[top]] < a[i]) --top;
    l[i] = stk[top] + 1;
    stk[++top] = i;
  }
  stk[top = 0] = n + 1;
  for (int i = (n); i >= (1); --i) {
    while (top > 0 && a[stk[top]] <= a[i]) --top;
    r[i] = stk[top] - 1;
    stk[++top] = i;
  }
  int ans = 0;
  for (int i = (1); i <= (n); ++i) {
    if (i - l[i] < r[i] - i) {
      for (int j = (l[i]); j <= (i); ++j) {
        int mn = std::max(i - j, K - 1), mx = r[i] - j + 1;
        (ans += (long long)((mx - 1) / (K - 1) - (mn - 1) / (K - 1)) * a[i] %
                mod) %= mod;
      }
    } else {
      for (int j = (r[i]); j >= (i); --j) {
        int mn = std::max(j - i, K - 1), mx = j - l[i] + 1;
        (ans += (long long)((mx - 1) / (K - 1) - (mn - 1) / (K - 1)) * a[i] %
                mod) %= mod;
      }
    }
  }
  cout << ans << '\n';
  return 0;
}

### 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 std::cin;
using std::cout;
const int maxn = 1000003, INF = 0x3f3f3f3f, mod = 1e9 + 7;
typedef int arr[maxn];
arr a, l, r, stk;
int main() {
  std::ios::sync_with_stdio(0);
  cin.tie(0);
  int n, K;
  cin >> n >> K;
  for (int i = (1); i <= (n); ++i) cin >> a[i];
  int top = 0;
  for (int i = (1); i <= (n); ++i) {
    while (top > 0 && a[stk[top]] < a[i]) --top;
    l[i] = stk[top] + 1;
    stk[++top] = i;
  }
  stk[top = 0] = n + 1;
  for (int i = (n); i >= (1); --i) {
    while (top > 0 && a[stk[top]] <= a[i]) --top;
    r[i] = stk[top] - 1;
    stk[++top] = i;
  }
  int ans = 0;
  for (int i = (1); i <= (n); ++i) {
    if (i - l[i] < r[i] - i) {
      for (int j = (l[i]); j <= (i); ++j) {
        int mn = std::max(i - j, K - 1), mx = r[i] - j + 1;
        (ans += (long long)((mx - 1) / (K - 1) - (mn - 1) / (K - 1)) * a[i] %
                mod) %= mod;
      }
    } else {
      for (int j = (r[i]); j >= (i); --j) {
        int mn = std::max(j - i, K - 1), mx = j - l[i] + 1;
        (ans += (long long)((mx - 1) / (K - 1) - (mn - 1) / (K - 1)) * a[i] %
                mod) %= mod;
      }
    }
  }
  cout << ans << '\n';
  return 0;
}

```