#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 50, mo = 1e9 + 7;
inline int gi() {
  char ch = getchar();
  int q = 0, x = 0;
  while (ch < '0' || ch > '9') q = (ch == '-' ? 1 : q), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return q ? (-x) : x;
}
int a[N], l[N], r[N], st[N], tot, ans;
int main() {
  int n = gi(), k = gi() - 1;
  for (int i = 1; i <= n; ++i) a[i] = gi();
  for (int i = 1; i <= n; ++i) {
    while (tot && a[i] > a[st[tot]]) r[st[tot--]] = i - 1;
    st[++tot] = i;
  }
  while (tot) r[st[tot--]] = n;
  for (int i = n; i >= 1; --i) {
    while (tot && a[i] >= a[st[tot]]) l[st[tot--]] = i + 1;
    st[++tot] = i;
  }
  while (tot) l[st[tot--]] = 1;
  for (int i = 1; i <= n; ++i) {
    int x = i - l[i], y = r[i] - i, u = x / k, v = y / k, z = (x + y) / k;
    if (x > y) swap(x, y), swap(u, v);
    int sum = ((u + 1ll) * u / 2 * k + u + (v - u) * (x + 1ll) +
               (z - v) * (x + y + 1ll) -
               ((z + 1ll) * z / 2 - (v + 1ll) * v / 2) * k) %
              mo * a[i] % mo;
    ((ans += sum) %= mo);
  }
  cout << ans;
  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 = 1e6 + 50, mo = 1e9 + 7;
inline int gi() {
  char ch = getchar();
  int q = 0, x = 0;
  while (ch < '0' || ch > '9') q = (ch == '-' ? 1 : q), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return q ? (-x) : x;
}
int a[N], l[N], r[N], st[N], tot, ans;
int main() {
  int n = gi(), k = gi() - 1;
  for (int i = 1; i <= n; ++i) a[i] = gi();
  for (int i = 1; i <= n; ++i) {
    while (tot && a[i] > a[st[tot]]) r[st[tot--]] = i - 1;
    st[++tot] = i;
  }
  while (tot) r[st[tot--]] = n;
  for (int i = n; i >= 1; --i) {
    while (tot && a[i] >= a[st[tot]]) l[st[tot--]] = i + 1;
    st[++tot] = i;
  }
  while (tot) l[st[tot--]] = 1;
  for (int i = 1; i <= n; ++i) {
    int x = i - l[i], y = r[i] - i, u = x / k, v = y / k, z = (x + y) / k;
    if (x > y) swap(x, y), swap(u, v);
    int sum = ((u + 1ll) * u / 2 * k + u + (v - u) * (x + 1ll) +
               (z - v) * (x + y + 1ll) -
               ((z + 1ll) * z / 2 - (v + 1ll) * v / 2) * k) %
              mo * a[i] % mo;
    ((ans += sum) %= mo);
  }
  cout << ans;
  return 0;
}

```