#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
void pd(int &x) {
  if (x >= 1000000007)
    x -= 1000000007;
  else if (x < 0)
    x += 1000000007;
}
int n, k, a[N];
int ans;
set<int> s;
set<int>::iterator it;
pair<int, int> rnk[N];
int f[N], len, l, r;
int solve(int l, int r, int mid) {
  r -= l - 1;
  mid -= l - 1;
  l = 1;
  int ans = f[r] - f[mid - 1];
  pd(ans);
  pd(ans -= f[r - mid]);
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pd(f[i] = f[i - 1] + (i - 1) / (k - 1));
  for (int i = 1; i <= n; i++) rnk[i] = make_pair(a[i], i);
  sort(rnk + 1, rnk + 1 + n);
  for (int i = n; i >= 1; i--) {
    int cur = rnk[i].second;
    s.insert(cur);
    it = s.find(cur);
    if (it != s.begin()) {
      it--;
      l = (*it) + 1;
      it++;
    } else
      l = 1;
    it++;
    if (it != s.end())
      r = (*it) - 1;
    else
      r = n;
    pd(ans += 1ll * solve(l, r, cur) * a[cur] % 1000000007);
  }
  cout << 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 = 1e6 + 10;
void pd(int &x) {
  if (x >= 1000000007)
    x -= 1000000007;
  else if (x < 0)
    x += 1000000007;
}
int n, k, a[N];
int ans;
set<int> s;
set<int>::iterator it;
pair<int, int> rnk[N];
int f[N], len, l, r;
int solve(int l, int r, int mid) {
  r -= l - 1;
  mid -= l - 1;
  l = 1;
  int ans = f[r] - f[mid - 1];
  pd(ans);
  pd(ans -= f[r - mid]);
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) pd(f[i] = f[i - 1] + (i - 1) / (k - 1));
  for (int i = 1; i <= n; i++) rnk[i] = make_pair(a[i], i);
  sort(rnk + 1, rnk + 1 + n);
  for (int i = n; i >= 1; i--) {
    int cur = rnk[i].second;
    s.insert(cur);
    it = s.find(cur);
    if (it != s.begin()) {
      it--;
      l = (*it) + 1;
      it++;
    } else
      l = 1;
    it++;
    if (it != s.end())
      r = (*it) - 1;
    else
      r = n;
    pd(ans += 1ll * solve(l, r, cur) * a[cur] % 1000000007);
  }
  cout << ans;
  return 0;
}

```