#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5, mod = 1e9 + 7;
int n, k, h, ans, a[N], st[N], l[N], r[N];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return f ? x : -x;
}
int calc(int x) {
  int y = (x - 1) / (k - 1);
  return (1ll * x * y - (1ll * (y + 1) * y >> 1) % mod * (k - 1)) % mod;
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    while (h && a[i] >= a[st[h]]) h--;
    l[i] = st[h], st[++h] = i;
  }
  st[h = 0] = n + 1;
  for (int i = n; i; i--) {
    while (h && a[i] > a[st[h]]) h--;
    r[i] = st[h], st[++h] = i;
  }
  for (int i = 1; i <= n; i++)
    ans = (ans + 1ll *
                     (calc(r[i] - l[i] - 1) - calc(i - l[i] - 1) -
                      calc(r[i] - i - 1)) *
                     a[i]) %
          mod;
  printf("%d", (ans + mod) % mod);
  return 0;
}

### Prompt

Your task is to create a CPP solution 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 + 5, mod = 1e9 + 7;
int n, k, h, ans, a[N], st[N], l[N], r[N];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) f ^= ch == '-', ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch ^ 48), ch = getchar();
  return f ? x : -x;
}
int calc(int x) {
  int y = (x - 1) / (k - 1);
  return (1ll * x * y - (1ll * (y + 1) * y >> 1) % mod * (k - 1)) % mod;
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    while (h && a[i] >= a[st[h]]) h--;
    l[i] = st[h], st[++h] = i;
  }
  st[h = 0] = n + 1;
  for (int i = n; i; i--) {
    while (h && a[i] > a[st[h]]) h--;
    r[i] = st[h], st[++h] = i;
  }
  for (int i = 1; i <= n; i++)
    ans = (ans + 1ll *
                     (calc(r[i] - l[i] - 1) - calc(i - l[i] - 1) -
                      calc(r[i] - i - 1)) *
                     a[i]) %
          mod;
  printf("%d", (ans + mod) % mod);
  return 0;
}

```