#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5, mod = 1e9 + 7;
int n, k, a[N], L[N], R[N], l, r, pr[N], ma;
int64_t ans, cnt;
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)
    for (L[i] = i - 1; L[i] >= 1 && a[i] >= a[L[i]]; L[i] = L[L[i]])
      ;
  for (int i = n; i >= 1; --i)
    for (R[i] = i + 1; R[i] <= n && a[i] > a[R[i]]; R[i] = R[R[i]])
      ;
  for (int i = 1; i <= n + k; ++i) pr[i] = i / (k - 1);
  for (int i = 1; i <= n; ++i) {
    l = L[i] + 1;
    r = R[i] - 1;
    cnt = 0;
    if (i - l < r - i)
      for (; l <= i; ++l) cnt += pr[r - l] - pr[i - l + k - 2] + 1;
    else
      for (; r >= i; --r) cnt += pr[r - l] - pr[r - i + k - 2] + 1;
    ans = (ans + (cnt - 1) % mod * a[i]) % mod;
    ma = max(ma, a[i]);
  }
  cout << ans;
}

### 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;
const int N = 2e6 + 5, mod = 1e9 + 7;
int n, k, a[N], L[N], R[N], l, r, pr[N], ma;
int64_t ans, cnt;
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)
    for (L[i] = i - 1; L[i] >= 1 && a[i] >= a[L[i]]; L[i] = L[L[i]])
      ;
  for (int i = n; i >= 1; --i)
    for (R[i] = i + 1; R[i] <= n && a[i] > a[R[i]]; R[i] = R[R[i]])
      ;
  for (int i = 1; i <= n + k; ++i) pr[i] = i / (k - 1);
  for (int i = 1; i <= n; ++i) {
    l = L[i] + 1;
    r = R[i] - 1;
    cnt = 0;
    if (i - l < r - i)
      for (; l <= i; ++l) cnt += pr[r - l] - pr[i - l + k - 2] + 1;
    else
      for (; r >= i; --r) cnt += pr[r - l] - pr[r - i + k - 2] + 1;
    ans = (ans + (cnt - 1) % mod * a[i]) % mod;
    ma = max(ma, a[i]);
  }
  cout << ans;
}

```