#include <bits/stdc++.h>
using namespace std;
const int N = 1000000 + 5;
const int mod = 1000000007;
int n, k;
int a[N], l[N], r[N];
int norm(long long x) {
  ((x %= mod) += mod) %= mod;
  return x;
}
int solve(int l, int r) {
  if (l > r || r - l + 1 < k) {
    return 0;
  }
  int p = r - l + 2 - k, f = p / (k - 1);
  return norm(1LL * p * (f + 1) % mod -
              1LL * (f + 1) * f / 2 % mod * (k - 1) % mod);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  stack<pair<int, int>> stk;
  for (int i = 1; i <= n; i++) {
    l[i] = i;
    while (!stk.empty() && stk.top().first < a[i]) {
      l[i] = stk.top().second;
      stk.pop();
    }
    stk.push(make_pair(a[i], l[i]));
  }
  while (!stk.empty()) {
    stk.pop();
  }
  for (int i = n; i >= 1; i--) {
    r[i] = i;
    while (!stk.empty() && stk.top().first <= a[i]) {
      r[i] = stk.top().second;
      stk.pop();
    }
    stk.push(make_pair(a[i], r[i]));
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int freq = norm(1LL * solve(l[i], r[i]) - 1LL * solve(l[i], i - 1) -
                    1LL * solve(i + 1, r[i]));
    ans = norm(ans + 1LL * freq * a[i]);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 = 1000000 + 5;
const int mod = 1000000007;
int n, k;
int a[N], l[N], r[N];
int norm(long long x) {
  ((x %= mod) += mod) %= mod;
  return x;
}
int solve(int l, int r) {
  if (l > r || r - l + 1 < k) {
    return 0;
  }
  int p = r - l + 2 - k, f = p / (k - 1);
  return norm(1LL * p * (f + 1) % mod -
              1LL * (f + 1) * f / 2 % mod * (k - 1) % mod);
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  stack<pair<int, int>> stk;
  for (int i = 1; i <= n; i++) {
    l[i] = i;
    while (!stk.empty() && stk.top().first < a[i]) {
      l[i] = stk.top().second;
      stk.pop();
    }
    stk.push(make_pair(a[i], l[i]));
  }
  while (!stk.empty()) {
    stk.pop();
  }
  for (int i = n; i >= 1; i--) {
    r[i] = i;
    while (!stk.empty() && stk.top().first <= a[i]) {
      r[i] = stk.top().second;
      stk.pop();
    }
    stk.push(make_pair(a[i], r[i]));
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int freq = norm(1LL * solve(l[i], r[i]) - 1LL * solve(l[i], i - 1) -
                    1LL * solve(i + 1, r[i]));
    ans = norm(ans + 1LL * freq * a[i]);
  }
  printf("%d\n", ans);
  return 0;
}

```