#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10, mod = 1e9 + 7;
int a[N], b[N], ans, w, l[N], r[N], ss[N];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int main() {
  int n = read(), k = read() - 1;
  for (int i = 1; i <= n; i++) a[i] = read();
  a[0] = a[n + 1] = 1e9 + 1;
  b[w = 1] = 0;
  for (int i = 1; i <= n; i++) {
    while (a[b[w]] <= a[i]) w--;
    l[i] = b[w];
    b[++w] = i;
  }
  b[w = 1] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (a[b[w]] < a[i]) w--;
    r[i] = b[w];
    b[++w] = i;
  }
  for (int i = 1; i * k < n; i++)
    ss[i] = ((long long)i * k + 1 + ss[i - 1]) % mod;
  for (int i = 1; i <= n; i++) {
    int x = i - l[i] - 1, y = r[i] - i - 1, z = a[i];
    if (x > y) swap(x, y);
    int s = x / k, p = y / k, q = (x + y) / k;
    ans = (((long long)ss[s] + (long long)(p - s) * (x + 1) +
            (long long)(x + y - k * (p + 1) + 1 + x + y - k * q + 1) * (q - p) /
                2) %
               mod * z +
           ans) %
          mod;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 + 10, mod = 1e9 + 7;
int a[N], b[N], ans, w, l[N], r[N], ss[N];
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int main() {
  int n = read(), k = read() - 1;
  for (int i = 1; i <= n; i++) a[i] = read();
  a[0] = a[n + 1] = 1e9 + 1;
  b[w = 1] = 0;
  for (int i = 1; i <= n; i++) {
    while (a[b[w]] <= a[i]) w--;
    l[i] = b[w];
    b[++w] = i;
  }
  b[w = 1] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (a[b[w]] < a[i]) w--;
    r[i] = b[w];
    b[++w] = i;
  }
  for (int i = 1; i * k < n; i++)
    ss[i] = ((long long)i * k + 1 + ss[i - 1]) % mod;
  for (int i = 1; i <= n; i++) {
    int x = i - l[i] - 1, y = r[i] - i - 1, z = a[i];
    if (x > y) swap(x, y);
    int s = x / k, p = y / k, q = (x + y) / k;
    ans = (((long long)ss[s] + (long long)(p - s) * (x + 1) +
            (long long)(x + y - k * (p + 1) + 1 + x + y - k * q + 1) * (q - p) /
                2) %
               mod * z +
           ans) %
          mod;
  }
  printf("%d\n", ans);
  return 0;
}

```