#include <bits/stdc++.h>
using namespace std;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) {
    v.emplace_back(x);
  }
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << ", ";
  err(++it, args...);
}
int a[1000006], L[1000006], R[1000006];
long long int counts[1000006];
void pre(int f, int d) {
  vector<int> v;
  vector<long long int> ps;
  while (f < 1000006 + d + d) {
    v.push_back(f);
    if (ps.size())
      ps.push_back(ps.back() + v.back());
    else
      ps.push_back(v.back());
    f += d;
  }
  int id = 0;
  for (int i = 1; i < 1000006; ++i) {
    if (v[id] == i) id++;
    counts[i] = 1LL * (i + 1) * id - (id > 0 ? ps[id - 1] : 0);
  }
}
int main() {
  int n, k, i;
  scanf("%d %d", &n, &k);
  pre(k, k - 1);
  for (i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  stack<int> st;
  for (i = 0; i < n; ++i) {
    while (st.size() and a[st.top()] <= a[i]) st.pop();
    if (st.size())
      L[i] = st.top() + 1;
    else
      L[i] = 0;
    st.push(i);
  }
  while (st.size()) st.pop();
  for (i = n - 1; i >= 0; --i) {
    while (st.size() and a[st.top()] < a[i]) st.pop();
    if (st.size())
      R[i] = st.top() - 1;
    else
      R[i] = n - 1;
    st.push(i);
  }
  long long int ans = 0;
  for (i = 0; i < n; ++i) {
    long long int x = counts[R[i] - L[i] + 1];
    if (R[i] - i >= 0) x -= counts[R[i] - i];
    if (i - L[i] >= 0) x -= counts[i - L[i]];
    x %= 1000000007;
    ans += 1LL * a[i] * x % 1000000007;
  }
  printf("%d", ans % 1000000007);
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) {
    v.emplace_back(x);
  }
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << ", ";
  err(++it, args...);
}
int a[1000006], L[1000006], R[1000006];
long long int counts[1000006];
void pre(int f, int d) {
  vector<int> v;
  vector<long long int> ps;
  while (f < 1000006 + d + d) {
    v.push_back(f);
    if (ps.size())
      ps.push_back(ps.back() + v.back());
    else
      ps.push_back(v.back());
    f += d;
  }
  int id = 0;
  for (int i = 1; i < 1000006; ++i) {
    if (v[id] == i) id++;
    counts[i] = 1LL * (i + 1) * id - (id > 0 ? ps[id - 1] : 0);
  }
}
int main() {
  int n, k, i;
  scanf("%d %d", &n, &k);
  pre(k, k - 1);
  for (i = 0; i < n; ++i) {
    scanf("%d", a + i);
  }
  stack<int> st;
  for (i = 0; i < n; ++i) {
    while (st.size() and a[st.top()] <= a[i]) st.pop();
    if (st.size())
      L[i] = st.top() + 1;
    else
      L[i] = 0;
    st.push(i);
  }
  while (st.size()) st.pop();
  for (i = n - 1; i >= 0; --i) {
    while (st.size() and a[st.top()] < a[i]) st.pop();
    if (st.size())
      R[i] = st.top() - 1;
    else
      R[i] = n - 1;
    st.push(i);
  }
  long long int ans = 0;
  for (i = 0; i < n; ++i) {
    long long int x = counts[R[i] - L[i] + 1];
    if (R[i] - i >= 0) x -= counts[R[i] - i];
    if (i - L[i] >= 0) x -= counts[i - L[i]];
    x %= 1000000007;
    ans += 1LL * a[i] * x % 1000000007;
  }
  printf("%d", ans % 1000000007);
}

```