#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const int N = 1000010;
const int oo = 1e9;
const int MOD = 1000000007;
int n, k;
long long a[N], sum = 0;
vector<long long> v[N];
deque<pair<long long, long long> > pr;
deque<long long> st;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  int idx = 1;
  for (int i = 1; i + k - 1 <= n; ++i) {
    while (idx < i + k) {
      while (st.size() && a[idx] > a[st.back()]) st.pop_back();
      st.push_back(idx);
      idx++;
    }
    while (st.size() && st.front() < i) st.pop_front();
    v[i % (k - 1)].push_back(a[st.front()]);
  }
  long long res = 0;
  for (int i = 0; i < k - 1; ++i) {
    sum = 0;
    pr.clear();
    for (int j = v[i].size() - 1; j >= 0; --j) {
      long long cur = 1;
      while (pr.size() && pr.front().first < v[i][j]) {
        cur += pr.front().second;
        sum -= pr.front().first * pr.front().second;
        pr.pop_front();
      }
      sum += cur * v[i][j];
      pr.push_front(make_pair(v[i][j], cur));
      res = (res + sum) % MOD;
    }
  }
  cout << res << endl;
  return 0;
}

### 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;
const double PI = acos(-1);
const int N = 1000010;
const int oo = 1e9;
const int MOD = 1000000007;
int n, k;
long long a[N], sum = 0;
vector<long long> v[N];
deque<pair<long long, long long> > pr;
deque<long long> st;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  int idx = 1;
  for (int i = 1; i + k - 1 <= n; ++i) {
    while (idx < i + k) {
      while (st.size() && a[idx] > a[st.back()]) st.pop_back();
      st.push_back(idx);
      idx++;
    }
    while (st.size() && st.front() < i) st.pop_front();
    v[i % (k - 1)].push_back(a[st.front()]);
  }
  long long res = 0;
  for (int i = 0; i < k - 1; ++i) {
    sum = 0;
    pr.clear();
    for (int j = v[i].size() - 1; j >= 0; --j) {
      long long cur = 1;
      while (pr.size() && pr.front().first < v[i][j]) {
        cur += pr.front().second;
        sum -= pr.front().first * pr.front().second;
        pr.pop_front();
      }
      sum += cur * v[i][j];
      pr.push_front(make_pair(v[i][j], cur));
      res = (res + sum) % MOD;
    }
  }
  cout << res << endl;
  return 0;
}

```