#include <bits/stdc++.h>
template <class T, class C = std::less<T>>
struct MaxQueue {
  MaxQueue() { clear(); }
  void clear() {
    sum = 0;
    id = 0;
    q.clear();
  }
  void push(T x) {
    std::pair<int, T> nxt(1, x);
    while (q.size() > id && cmp(q.back().second, x)) {
      nxt.first += q.back().first;
      sum -= (long long)q.back().first * q.back().second;
      q.pop_back();
    }
    q.push_back(nxt);
    sum += (long long)q.back().first * q.back().second;
  }
  T qry() { return q[id].second; }
  long long getSum() { return sum; }
  void pop() {
    q[id].first--;
    if (q[id].first == 0) {
      id++;
    }
  }

 private:
  std::vector<std::pair<int, T>> q;
  long long sum;
  int id;
  C cmp;
};
const int ms = 2001000;
const int MOD = 1e9 + 7;
int a[ms];
int main() {
  int n, k;
  std::cin >> n >> k;
  MaxQueue<int> q;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    q.push(a[i]);
    if (i >= k - 1) {
      a[i - (k - 1)] = q.qry();
      q.pop();
    }
  }
  k--;
  int ans = 0;
  std::vector<MaxQueue<int>> v(k);
  for (int i = n - 1; i >= 0; i--) {
    v[i % k].push(a[i]);
    ans = (int)(ans + v[i % k].getSum() % MOD) % MOD;
    ans = (int)(ans - v[i % k].qry() % MOD) % MOD;
    ans = (ans + MOD) % MOD;
  }
  std::cout << ans << std::endl;
}

### Prompt

In Cpp, your task is to solve 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>
template <class T, class C = std::less<T>>
struct MaxQueue {
  MaxQueue() { clear(); }
  void clear() {
    sum = 0;
    id = 0;
    q.clear();
  }
  void push(T x) {
    std::pair<int, T> nxt(1, x);
    while (q.size() > id && cmp(q.back().second, x)) {
      nxt.first += q.back().first;
      sum -= (long long)q.back().first * q.back().second;
      q.pop_back();
    }
    q.push_back(nxt);
    sum += (long long)q.back().first * q.back().second;
  }
  T qry() { return q[id].second; }
  long long getSum() { return sum; }
  void pop() {
    q[id].first--;
    if (q[id].first == 0) {
      id++;
    }
  }

 private:
  std::vector<std::pair<int, T>> q;
  long long sum;
  int id;
  C cmp;
};
const int ms = 2001000;
const int MOD = 1e9 + 7;
int a[ms];
int main() {
  int n, k;
  std::cin >> n >> k;
  MaxQueue<int> q;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    q.push(a[i]);
    if (i >= k - 1) {
      a[i - (k - 1)] = q.qry();
      q.pop();
    }
  }
  k--;
  int ans = 0;
  std::vector<MaxQueue<int>> v(k);
  for (int i = n - 1; i >= 0; i--) {
    v[i % k].push(a[i]);
    ans = (int)(ans + v[i % k].getSum() % MOD) % MOD;
    ans = (int)(ans - v[i % k].qry() % MOD) % MOD;
    ans = (ans + MOD) % MOD;
  }
  std::cout << ans << std::endl;
}

```