#include <bits/stdc++.h>
using namespace std;
const int p = 1e9 + 7;
int n, k;
int L[1001000], R[1001000];
int a[1001000];
int ans;
int read() {
  int sum = 0;
  char c = getchar();
  bool flag = true;
  while (c < '0' || c > '9') {
    if (c == '-') flag = false;
    c = getchar();
  }
  while (c >= '0' && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  if (flag)
    return sum;
  else
    return -sum;
}
stack<int> q;
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n; ++i) {
    while (!q.empty() && a[i] > a[q.top()]) q.pop();
    if (q.empty())
      L[i] = 1;
    else
      L[i] = q.top() + 1;
    q.push(i);
  }
  while (!q.empty()) q.pop();
  for (int i = n; i >= 1; --i) {
    while (!q.empty() && a[i] >= a[q.top()]) q.pop();
    if (q.empty())
      R[i] = n;
    else
      R[i] = q.top() - 1;
    q.push(i);
  }
  k--;
  for (int i = 1; i <= n; ++i) {
    int x = i - L[i], y = R[i] - i;
    if (x > y) swap(x, y);
    int N = x / k, M = y / k, H = (x + y) / k;
    ans = ans + (1ll * N * (N + 1) / 2 * k % p + N) * a[i] % p;
    ans %= p;
    ans = ans + (1ll * (x + 1) * (M - N) % p) * a[i] % p;
    ans %= p;
    ans = ans + (1ll * (x + y + 1ll) * (H - M) % p) * a[i] % p;
    ans %= p;
    ans = ans - ((H - M) * (H + M + 1LL) / 2 * k % p) * a[i] % p;
    ans %= p;
    ans %= p;
  }
  ans = (ans + p) % p;
  printf("%d\n", ans);
}

### Prompt

Please provide a Cpp coded solution 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 p = 1e9 + 7;
int n, k;
int L[1001000], R[1001000];
int a[1001000];
int ans;
int read() {
  int sum = 0;
  char c = getchar();
  bool flag = true;
  while (c < '0' || c > '9') {
    if (c == '-') flag = false;
    c = getchar();
  }
  while (c >= '0' && c <= '9') sum = sum * 10 + c - 48, c = getchar();
  if (flag)
    return sum;
  else
    return -sum;
}
stack<int> q;
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  for (int i = 1; i <= n; ++i) {
    while (!q.empty() && a[i] > a[q.top()]) q.pop();
    if (q.empty())
      L[i] = 1;
    else
      L[i] = q.top() + 1;
    q.push(i);
  }
  while (!q.empty()) q.pop();
  for (int i = n; i >= 1; --i) {
    while (!q.empty() && a[i] >= a[q.top()]) q.pop();
    if (q.empty())
      R[i] = n;
    else
      R[i] = q.top() - 1;
    q.push(i);
  }
  k--;
  for (int i = 1; i <= n; ++i) {
    int x = i - L[i], y = R[i] - i;
    if (x > y) swap(x, y);
    int N = x / k, M = y / k, H = (x + y) / k;
    ans = ans + (1ll * N * (N + 1) / 2 * k % p + N) * a[i] % p;
    ans %= p;
    ans = ans + (1ll * (x + 1) * (M - N) % p) * a[i] % p;
    ans %= p;
    ans = ans + (1ll * (x + y + 1ll) * (H - M) % p) * a[i] % p;
    ans %= p;
    ans = ans - ((H - M) * (H + M + 1LL) / 2 * k % p) * a[i] % p;
    ans %= p;
    ans %= p;
  }
  ans = (ans + p) % p;
  printf("%d\n", ans);
}

```