#include <bits/stdc++.h>
using namespace std;
long long a[1000002];
long long mod = 1e9 + 7;
long long r[1000002];
long long l[1000002];
long long summa[1000002];
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  stack<long long> st;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (i > 0) {
      summa[i + 1] = (summa[i] + (i) / (k - 1)) % mod;
    }
  }
  for (int i = 0; i < n; ++i) {
    while (!st.empty() && a[i] > a[st.top()]) {
      r[st.top()] = i - 1;
      st.pop();
    }
    st.push(i);
  }
  while (!st.empty()) {
    r[st.top()] = n - 1;
    st.pop();
  }
  for (int i = n - 1; i >= 0; --i) {
    while (!st.empty() && a[i] >= a[st.top()]) {
      l[st.top()] = i + 1;
      st.pop();
    }
    st.push(i);
  }
  while (!st.empty()) {
    l[st.top()] = 0;
    st.pop();
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    long long x =
        (summa[r[i] - l[i] + 1] - summa[i - l[i]] - summa[r[i] - i] + 2 * mod) %
        mod;
    ans = (ans + x * a[i]) % mod;
  }
  cout << ans << "\n";
  return 0;
}

### Prompt

Please create a solution in cpp 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;
long long a[1000002];
long long mod = 1e9 + 7;
long long r[1000002];
long long l[1000002];
long long summa[1000002];
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL), cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  stack<long long> st;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    if (i > 0) {
      summa[i + 1] = (summa[i] + (i) / (k - 1)) % mod;
    }
  }
  for (int i = 0; i < n; ++i) {
    while (!st.empty() && a[i] > a[st.top()]) {
      r[st.top()] = i - 1;
      st.pop();
    }
    st.push(i);
  }
  while (!st.empty()) {
    r[st.top()] = n - 1;
    st.pop();
  }
  for (int i = n - 1; i >= 0; --i) {
    while (!st.empty() && a[i] >= a[st.top()]) {
      l[st.top()] = i + 1;
      st.pop();
    }
    st.push(i);
  }
  while (!st.empty()) {
    l[st.top()] = 0;
    st.pop();
  }
  long long ans = 0;
  for (int i = 0; i < n; ++i) {
    long long x =
        (summa[r[i] - l[i] + 1] - summa[i - l[i]] - summa[r[i] - i] + 2 * mod) %
        mod;
    ans = (ans + x * a[i]) % mod;
  }
  cout << ans << "\n";
  return 0;
}

```