#include <bits/stdc++.h>
using namespace std;
template <class I>
inline void ckMax(I& p, I q) {
  p = (p > q ? p : q);
}
template <class I>
inline void ckMin(I& p, I q) {
  p = (p < q ? p : q);
}
const int N = 1000005;
int n, m, a[N], l[N], r[N], s[N], top, ans, k;
inline int cal(const int len) {
  const int x = len / k,
            y = ((0ll + len) * x - ((1ll + x) * x >> 1) % 1000000007 * k) %
                1000000007;
  return y < 0 ? y + 1000000007 : y;
}
int main() {
  scanf("%d%d", &n, &k), --k;
  register int i;
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (i = 1; i <= n; ++i) {
    while (top && a[s[top]] <= a[i]) --top;
    l[i] = s[top] + 1, s[++top] = i;
  }
  for (i = n, s[top = 0] = n + 1; i; --i) {
    while (top && a[s[top]] < a[i]) --top;
    r[i] = s[top] - 1, s[++top] = i;
  }
  for (i = 1; i <= n; ++i)
    ans = (ans + (0ll + cal(r[i] - l[i] + 1) + 1000000007 - cal(i - l[i]) +
                  1000000007 - cal(r[i] - i)) *
                     a[i]) %
          1000000007;
  printf("%d", ans < 0 ? ans + 1000000007 : ans);
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
template <class I>
inline void ckMax(I& p, I q) {
  p = (p > q ? p : q);
}
template <class I>
inline void ckMin(I& p, I q) {
  p = (p < q ? p : q);
}
const int N = 1000005;
int n, m, a[N], l[N], r[N], s[N], top, ans, k;
inline int cal(const int len) {
  const int x = len / k,
            y = ((0ll + len) * x - ((1ll + x) * x >> 1) % 1000000007 * k) %
                1000000007;
  return y < 0 ? y + 1000000007 : y;
}
int main() {
  scanf("%d%d", &n, &k), --k;
  register int i;
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (i = 1; i <= n; ++i) {
    while (top && a[s[top]] <= a[i]) --top;
    l[i] = s[top] + 1, s[++top] = i;
  }
  for (i = n, s[top = 0] = n + 1; i; --i) {
    while (top && a[s[top]] < a[i]) --top;
    r[i] = s[top] - 1, s[++top] = i;
  }
  for (i = 1; i <= n; ++i)
    ans = (ans + (0ll + cal(r[i] - l[i] + 1) + 1000000007 - cal(i - l[i]) +
                  1000000007 - cal(r[i] - i)) *
                     a[i]) %
          1000000007;
  printf("%d", ans < 0 ? ans + 1000000007 : ans);
  return 0;
}

```