#include <bits/stdc++.h>
using namespace std;
int n, k, a[1000020], prv[1000020], nxt[1000020];
stack<int> st;
const int mod = 1e9 + 7;
inline int func(int l) {
  int t = l / (k - 1);
  return ((1ll * l * t - 1ll * t * (t + 1) / 2 * (k - 1)) % mod + mod) % mod;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  while (!st.empty()) st.pop();
  st.push(0);
  for (int i = 1; i <= n; i++) {
    while (st.top() && a[st.top()] < a[i]) st.pop();
    prv[i] = st.top(), st.push(i);
  }
  while (!st.empty()) st.pop();
  st.push(n + 1);
  for (int i = n; i; i--) {
    while (st.top() <= n && a[st.top()] <= a[i]) st.pop();
    nxt[i] = st.top(), st.push(i);
  }
  int rlt = 0;
  for (int i = 1; i <= n; i++) {
    rlt = (rlt + (2ll * mod + func(nxt[i] - prv[i] - 1) - func(nxt[i] - i - 1) -
                  func(i - prv[i] - 1)) %
                     mod * a[i]) %
          mod;
  }
  printf("%d\n", rlt);
}

### Prompt

Create a solution in Cpp for 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, a[1000020], prv[1000020], nxt[1000020];
stack<int> st;
const int mod = 1e9 + 7;
inline int func(int l) {
  int t = l / (k - 1);
  return ((1ll * l * t - 1ll * t * (t + 1) / 2 * (k - 1)) % mod + mod) % mod;
}
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  while (!st.empty()) st.pop();
  st.push(0);
  for (int i = 1; i <= n; i++) {
    while (st.top() && a[st.top()] < a[i]) st.pop();
    prv[i] = st.top(), st.push(i);
  }
  while (!st.empty()) st.pop();
  st.push(n + 1);
  for (int i = n; i; i--) {
    while (st.top() <= n && a[st.top()] <= a[i]) st.pop();
    nxt[i] = st.top(), st.push(i);
  }
  int rlt = 0;
  for (int i = 1; i <= n; i++) {
    rlt = (rlt + (2ll * mod + func(nxt[i] - prv[i] - 1) - func(nxt[i] - i - 1) -
                  func(i - prv[i] - 1)) %
                     mod * a[i]) %
          mod;
  }
  printf("%d\n", rlt);
}

```