#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
inline void add(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int n, k, a[1001005];
pair<int, int> p[1001005];
set<int> st;
inline int calc(int len, int k) {
  int z = len / (k - 1);
  return (z * (int64_t)len - z * int64_t(z + 1) / 2 * (k - 1)) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    p[i] = {a[i], i + 1};
  }
  st.insert(0);
  st.insert(n + 1);
  sort(p, p + n, greater<pair<int, int> >());
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    int idx = p[i].second;
    auto it = st.lower_bound(idx);
    int R = *it - idx - 1;
    int L = idx - *--it - 1;
    st.insert(idx);
    int cur = calc(L + R + 1, k);
    add(cur, mod - calc(L, k));
    add(cur, mod - calc(R, k));
    add(ans, cur * int64_t(p[i].first) % mod);
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in cpp 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 mod = 1000000007;
inline void add(int& x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int n, k, a[1001005];
pair<int, int> p[1001005];
set<int> st;
inline int calc(int len, int k) {
  int z = len / (k - 1);
  return (z * (int64_t)len - z * int64_t(z + 1) / 2 * (k - 1)) % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    p[i] = {a[i], i + 1};
  }
  st.insert(0);
  st.insert(n + 1);
  sort(p, p + n, greater<pair<int, int> >());
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    int idx = p[i].second;
    auto it = st.lower_bound(idx);
    int R = *it - idx - 1;
    int L = idx - *--it - 1;
    st.insert(idx);
    int cur = calc(L + R + 1, k);
    add(cur, mod - calc(L, k));
    add(cur, mod - calc(R, k));
    add(ans, cur * int64_t(p[i].first) % mod);
  }
  cout << ans << endl;
}

```