#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long cnt(int len, int k) {
  long long n = len / (k - 1);
  long long ans = n * len - (n * (n + 1)) * (k - 1) / 2;
  return ans % mod;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  stack<int> st;
  vector<int> left(n + 5), right(n + 5);
  for (int i = 0; i < n; i++) {
    while (!st.empty() && a[st.top()] <= a[i]) {
      st.pop();
    }
    if (st.empty()) {
      left[i] = 0;
    } else {
      left[i] = st.top() + 1;
    }
    st.push(i);
  }
  while (!st.empty()) {
    st.pop();
  }
  for (int i = n - 1; i >= 0; i--) {
    while (!st.empty() && a[st.top()] < a[i]) {
      st.pop();
    }
    if (st.empty()) {
      right[i] = n - 1;
    } else {
      right[i] = st.top() - 1;
    }
    st.push(i);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long tot = right[i] - left[i] + 1;
    long long ex1 = i - left[i];
    long long ex2 = right[i] - i;
    long long countit = cnt(tot, k) - cnt(ex1, k) - cnt(ex2, k);
    countit %= mod;
    countit = (a[i] * countit) % mod;
    ans = (ans + countit) % mod;
  }
  cout << ans % mod << '\n';
}

### 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>
using namespace std;
long long mod = 1e9 + 7;
long long cnt(int len, int k) {
  long long n = len / (k - 1);
  long long ans = n * len - (n * (n + 1)) * (k - 1) / 2;
  return ans % mod;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  stack<int> st;
  vector<int> left(n + 5), right(n + 5);
  for (int i = 0; i < n; i++) {
    while (!st.empty() && a[st.top()] <= a[i]) {
      st.pop();
    }
    if (st.empty()) {
      left[i] = 0;
    } else {
      left[i] = st.top() + 1;
    }
    st.push(i);
  }
  while (!st.empty()) {
    st.pop();
  }
  for (int i = n - 1; i >= 0; i--) {
    while (!st.empty() && a[st.top()] < a[i]) {
      st.pop();
    }
    if (st.empty()) {
      right[i] = n - 1;
    } else {
      right[i] = st.top() - 1;
    }
    st.push(i);
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long tot = right[i] - left[i] + 1;
    long long ex1 = i - left[i];
    long long ex2 = right[i] - i;
    long long countit = cnt(tot, k) - cnt(ex1, k) - cnt(ex2, k);
    countit %= mod;
    countit = (a[i] * countit) % mod;
    ans = (ans + countit) % mod;
  }
  cout << ans % mod << '\n';
}

```