#include <bits/stdc++.h>
using namespace std;
bool comp(pair<long long, long long> p1, pair<long long, long long> p2) {
  return p1.second < p2.second;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  k--;
  vector<long long> v(n + 1);
  vector<long long> left(n + 1);
  vector<long long> right(n + 1);
  stack<long long> st;
  for (long long i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (long long i = 1; i <= n; i++) {
    while (!st.empty() && v[st.top()] < v[i]) {
      right[st.top()] = i - 1;
      st.pop();
    }
    st.push(i);
  }
  while (!st.empty()) {
    right[st.top()] = n;
    st.pop();
  }
  for (long long i = n; i >= 1; i--) {
    while (!st.empty() && v[st.top()] <= v[i]) {
      left[st.top()] = i + 1;
      st.pop();
    }
    st.push(i);
  }
  while (!st.empty()) {
    left[st.top()] = 1;
    st.pop();
  }
  vector<long long> cl(n + 1), fl(n + 1);
  vector<long long> ccl(n + 1, 0), cfl(n + 1, 0);
  long long csf = 0, csc = 0;
  for (long long i = 1; i <= n; i++) {
    csf += i / k;
    fl[i] = csf;
    csc += (i + k - 1) / k;
    cl[i] = csc;
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    long long l1 = i - left[i];
    long long l2 = max((long long)1, i - right[i] + k);
    long long lans = 0;
    if (l2 <= l1) {
      lans -= (cl[l1] - cl[l2 - 1]);
      long long r1 = right[i] - left[i];
      long long r2 = right[i] - i + 1;
      lans += (fl[r1] - fl[r2 - 1]);
      lans += l1 - l2 + 1;
      lans = lans % 1000000007;
      ans += lans * v[i];
      ans = ans % 1000000007;
    }
    ans += ((right[i] - i) / (k)) * v[i];
    ans = ans % 1000000007;
  }
  cout << ans;
  return 0;
}

### Prompt

Your challenge is to write 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;
bool comp(pair<long long, long long> p1, pair<long long, long long> p2) {
  return p1.second < p2.second;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  k--;
  vector<long long> v(n + 1);
  vector<long long> left(n + 1);
  vector<long long> right(n + 1);
  stack<long long> st;
  for (long long i = 1; i <= n; i++) {
    cin >> v[i];
  }
  for (long long i = 1; i <= n; i++) {
    while (!st.empty() && v[st.top()] < v[i]) {
      right[st.top()] = i - 1;
      st.pop();
    }
    st.push(i);
  }
  while (!st.empty()) {
    right[st.top()] = n;
    st.pop();
  }
  for (long long i = n; i >= 1; i--) {
    while (!st.empty() && v[st.top()] <= v[i]) {
      left[st.top()] = i + 1;
      st.pop();
    }
    st.push(i);
  }
  while (!st.empty()) {
    left[st.top()] = 1;
    st.pop();
  }
  vector<long long> cl(n + 1), fl(n + 1);
  vector<long long> ccl(n + 1, 0), cfl(n + 1, 0);
  long long csf = 0, csc = 0;
  for (long long i = 1; i <= n; i++) {
    csf += i / k;
    fl[i] = csf;
    csc += (i + k - 1) / k;
    cl[i] = csc;
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    long long l1 = i - left[i];
    long long l2 = max((long long)1, i - right[i] + k);
    long long lans = 0;
    if (l2 <= l1) {
      lans -= (cl[l1] - cl[l2 - 1]);
      long long r1 = right[i] - left[i];
      long long r2 = right[i] - i + 1;
      lans += (fl[r1] - fl[r2 - 1]);
      lans += l1 - l2 + 1;
      lans = lans % 1000000007;
      ans += lans * v[i];
      ans = ans % 1000000007;
    }
    ans += ((right[i] - i) / (k)) * v[i];
    ans = ans % 1000000007;
  }
  cout << ans;
  return 0;
}

```