#include <bits/stdc++.h>
using namespace std;
int const N = 1000000, md = 1e9 + 7;
int sg[N + 1], x[N + 2], l[N + 1], r[N + 1];
inline void ad(int &x, int y) {
  if ((x += y) >= md) x -= md;
}
inline void sb(int &x, int y) {
  if ((x -= y) < 0) x += md;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < (int)(n + 1); ++i) scanf("%d", x + i);
  int s = 0;
  for (int i = k; i < (int)(n + 1); ++i) {
    if ((i - 1) % (k - 1) == 0) ++s;
    sg[i] = sg[i - 1];
    ad(sg[i], s);
  }
  x[0] = 2e9;
  x[n + 1] = 2e9;
  stack<int> st;
  st.push(0);
  for (int i = 1; i < (int)(n + 1); ++i) {
    while (x[i] > x[st.top()]) st.pop();
    l[i] = st.top() + 1;
    st.push(i);
  }
  while (!st.empty()) st.pop();
  st.push(n + 1);
  for (int i = n; i > 0; --i) {
    while (x[i] >= x[st.top()]) st.pop();
    r[i] = st.top() - 1;
    st.push(i);
  }
  int an = 0;
  for (int i = 1; i < (int)(n + 1); ++i) {
    int z = sg[r[i] - l[i] + 1];
    sb(z, sg[i - l[i]]);
    sb(z, sg[r[i] - i]);
    ad(an, (long long)z * x[i] % md);
  }
  printf("%d\n", an);
}

### Prompt

Please formulate 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;
int const N = 1000000, md = 1e9 + 7;
int sg[N + 1], x[N + 2], l[N + 1], r[N + 1];
inline void ad(int &x, int y) {
  if ((x += y) >= md) x -= md;
}
inline void sb(int &x, int y) {
  if ((x -= y) < 0) x += md;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < (int)(n + 1); ++i) scanf("%d", x + i);
  int s = 0;
  for (int i = k; i < (int)(n + 1); ++i) {
    if ((i - 1) % (k - 1) == 0) ++s;
    sg[i] = sg[i - 1];
    ad(sg[i], s);
  }
  x[0] = 2e9;
  x[n + 1] = 2e9;
  stack<int> st;
  st.push(0);
  for (int i = 1; i < (int)(n + 1); ++i) {
    while (x[i] > x[st.top()]) st.pop();
    l[i] = st.top() + 1;
    st.push(i);
  }
  while (!st.empty()) st.pop();
  st.push(n + 1);
  for (int i = n; i > 0; --i) {
    while (x[i] >= x[st.top()]) st.pop();
    r[i] = st.top() - 1;
    st.push(i);
  }
  int an = 0;
  for (int i = 1; i < (int)(n + 1); ++i) {
    int z = sg[r[i] - l[i] + 1];
    sb(z, sg[i - l[i]]);
    sb(z, sg[r[i] - i]);
    ad(an, (long long)z * x[i] % md);
  }
  printf("%d\n", an);
}

```