#include <bits/stdc++.h>
using namespace std;
int n, k;
int a[1000007];
int l[1000007];
int r[1000007];
long long f[1000007];
void input() {
  scanf("%d%d", &n, &k);
  int i;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
}
void solve() {
  stack<int> s;
  int i;
  for (i = 1; i <= n; i++) {
    while (s.empty() == false && a[s.top()] < a[i]) {
      s.pop();
    }
    if (s.empty() == true) {
      l[i] = 0;
    } else {
      l[i] = s.top();
    }
    s.push(i);
  }
  while (s.empty() == false) {
    s.pop();
  }
  for (i = n; i >= 1; i--) {
    while (s.empty() == false && a[s.top()] <= a[i]) {
      s.pop();
    }
    if (s.empty() == true) {
      r[i] = (n + 1);
    } else {
      r[i] = s.top();
    }
    s.push(i);
  }
  for (i = 1; i <= n; i++) {
    if (i < k) {
      f[i] = 0;
      continue;
    }
    f[i] = (i - k + 1);
    long long val = (i - k) / (k - 1);
    long long aux = ((val * (i - k + 1)) % 1000000007) -
                    (((val * (val + 1) / 2) * (k - 1)) % 1000000007);
    if (aux < 0) {
      aux += 1000000007;
    }
    f[i] = (f[i] + aux) % 1000000007;
  }
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    int x = i - l[i] - 1;
    int y = r[i] - i - 1;
    long long cur =
        (f[x + y + 1] - ((f[x] + f[y]) % 1000000007) + 1000000007) % 1000000007;
    ans += (cur * a[i]) % 1000000007;
    ans %= 1000000007;
  }
  printf("%I64d\n", ans);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int n, k;
int a[1000007];
int l[1000007];
int r[1000007];
long long f[1000007];
void input() {
  scanf("%d%d", &n, &k);
  int i;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
}
void solve() {
  stack<int> s;
  int i;
  for (i = 1; i <= n; i++) {
    while (s.empty() == false && a[s.top()] < a[i]) {
      s.pop();
    }
    if (s.empty() == true) {
      l[i] = 0;
    } else {
      l[i] = s.top();
    }
    s.push(i);
  }
  while (s.empty() == false) {
    s.pop();
  }
  for (i = n; i >= 1; i--) {
    while (s.empty() == false && a[s.top()] <= a[i]) {
      s.pop();
    }
    if (s.empty() == true) {
      r[i] = (n + 1);
    } else {
      r[i] = s.top();
    }
    s.push(i);
  }
  for (i = 1; i <= n; i++) {
    if (i < k) {
      f[i] = 0;
      continue;
    }
    f[i] = (i - k + 1);
    long long val = (i - k) / (k - 1);
    long long aux = ((val * (i - k + 1)) % 1000000007) -
                    (((val * (val + 1) / 2) * (k - 1)) % 1000000007);
    if (aux < 0) {
      aux += 1000000007;
    }
    f[i] = (f[i] + aux) % 1000000007;
  }
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    int x = i - l[i] - 1;
    int y = r[i] - i - 1;
    long long cur =
        (f[x + y + 1] - ((f[x] + f[y]) % 1000000007) + 1000000007) % 1000000007;
    ans += (cur * a[i]) % 1000000007;
    ans %= 1000000007;
  }
  printf("%I64d\n", ans);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  input();
  solve();
  return 0;
}

```