#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
const long long mod = 1e9 + 7;
long long f[N];
struct pos {
  int x;
  int p;
} a[N];
bool cmp(pos a, pos b) { return a.x > b.x; }
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (long long i = k; i <= n; i++) f[i] = (f[i + 1 - k] + i - k + 1) % mod;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i].p = i;
  }
  set<int> s;
  s.insert(0), s.insert(n + 1);
  long long ans = 0;
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    auto it = s.lower_bound(a[i].p);
    int r = *it, l = *(--it);
    s.insert(a[i].p);
    long long now =
        ((f[r - l - 1] - f[r - a[i].p - 1] - f[a[i].p - l - 1]) % mod + mod) %
        mod;
    ans = (ans + (now * a[i].x) % mod) % mod;
  }
  cout << ans % mod << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp to 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 = 1000005;
const long long mod = 1e9 + 7;
long long f[N];
struct pos {
  int x;
  int p;
} a[N];
bool cmp(pos a, pos b) { return a.x > b.x; }
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (long long i = k; i <= n; i++) f[i] = (f[i + 1 - k] + i - k + 1) % mod;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    a[i].p = i;
  }
  set<int> s;
  s.insert(0), s.insert(n + 1);
  long long ans = 0;
  sort(a + 1, a + n + 1, cmp);
  for (int i = 1; i <= n; i++) {
    auto it = s.lower_bound(a[i].p);
    int r = *it, l = *(--it);
    s.insert(a[i].p);
    long long now =
        ((f[r - l - 1] - f[r - a[i].p - 1] - f[a[i].p - l - 1]) % mod + mod) %
        mod;
    ans = (ans + (now * a[i].x) % mod) % mod;
  }
  cout << ans % mod << endl;
  return 0;
}

```