#include <bits/stdc++.h>
const int N = 1e6 + 5, mo = 1e9 + 7;
int n, k, a[N], i, lb[N], rb[N], st[N], w, ans;
inline long long calc(int x, int y) {
  if (k == 1) return 1ll * x * y - 1;
  int xx, yy;
  xx = x / k;
  xx += x % k != 0;
  yy = y / k;
  yy += y % k != 0;
  long long ans = 1ll * xx * yy - 1 + 1ll * (x - xx) * (y / k);
  xx = x % k;
  yy = y % k;
  if (xx > 1 && yy > 1) ans += std::max(0, xx - std::max(1, (k - yy + 1)));
  if (yy > 1) ans += 1ll * (x / k) * std::max(0, k - std::max(1, (k - yy + 1)));
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (i = 1; i <= n; ++i) scanf("%d", a + i);
  for (i = 1; i <= n; ++i) {
    for (; w && a[st[w]] < a[i]; --w)
      ;
    lb[i] = st[w];
    st[++w] = i;
  }
  st[0] = n + 1;
  w = 0;
  for (i = n; i; --i) {
    for (; w && a[st[w]] <= a[i]; --w)
      ;
    rb[i] = st[w];
    st[++w] = i;
  }
  for (i = 1; i <= n; ++i)
    ans = (ans + std::max(0ll, calc(i - lb[i], rb[i] - i)) % mo * a[i]) % mo;
  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>
const int N = 1e6 + 5, mo = 1e9 + 7;
int n, k, a[N], i, lb[N], rb[N], st[N], w, ans;
inline long long calc(int x, int y) {
  if (k == 1) return 1ll * x * y - 1;
  int xx, yy;
  xx = x / k;
  xx += x % k != 0;
  yy = y / k;
  yy += y % k != 0;
  long long ans = 1ll * xx * yy - 1 + 1ll * (x - xx) * (y / k);
  xx = x % k;
  yy = y % k;
  if (xx > 1 && yy > 1) ans += std::max(0, xx - std::max(1, (k - yy + 1)));
  if (yy > 1) ans += 1ll * (x / k) * std::max(0, k - std::max(1, (k - yy + 1)));
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (i = 1; i <= n; ++i) scanf("%d", a + i);
  for (i = 1; i <= n; ++i) {
    for (; w && a[st[w]] < a[i]; --w)
      ;
    lb[i] = st[w];
    st[++w] = i;
  }
  st[0] = n + 1;
  w = 0;
  for (i = n; i; --i) {
    for (; w && a[st[w]] <= a[i]; --w)
      ;
    rb[i] = st[w];
    st[++w] = i;
  }
  for (i = 1; i <= n; ++i)
    ans = (ans + std::max(0ll, calc(i - lb[i], rb[i] - i)) % mo * a[i]) % mo;
  printf("%d\n", ans);
  return 0;
}

```