#include <bits/stdc++.h>
using namespace std;
long long int r[1000005], l[1000005];
long long int A[1000005];
long long int dp[1000005 + 1] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> A[i];
  r[n - 1] = 0;
  stack<int> s;
  s.push(n - 1);
  for (int i = n - 2; i >= 0; i--) {
    while (!s.empty() && A[s.top()] < A[i]) s.pop();
    if (!s.empty())
      r[i] = s.top() - i - 1;
    else
      r[i] = n - i - 1;
    s.push(i);
  }
  while (!s.empty()) s.pop();
  l[0] = 0;
  s.push(0);
  for (int i = 1; i < n; i++) {
    while (!s.empty() && A[s.top()] <= A[i]) s.pop();
    if (!s.empty())
      l[i] = i - s.top() - 1;
    else
      l[i] = i;
    s.push(i);
  }
  for (int i = k; i <= n; i += (k - 1)) dp[i] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i] += dp[i - 1];
  }
  for (int i = 1; i <= n; i++) dp[i] = (dp[i] + dp[i - 1]) % 1000000007;
  long long int sum = 0;
  long long int temp;
  for (int i = 0; i < n; i++) {
    temp = (dp[l[i] + r[i] + 1] - dp[r[i]] - dp[l[i]]) % 1000000007;
    if (temp < 0) temp += 1000000007;
    sum = (sum + 1LL * temp * A[i]) % 1000000007;
  }
  sum = sum % 1000000007;
  cout << sum << "\n";
}

### 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;
long long int r[1000005], l[1000005];
long long int A[1000005];
long long int dp[1000005 + 1] = {0};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> A[i];
  r[n - 1] = 0;
  stack<int> s;
  s.push(n - 1);
  for (int i = n - 2; i >= 0; i--) {
    while (!s.empty() && A[s.top()] < A[i]) s.pop();
    if (!s.empty())
      r[i] = s.top() - i - 1;
    else
      r[i] = n - i - 1;
    s.push(i);
  }
  while (!s.empty()) s.pop();
  l[0] = 0;
  s.push(0);
  for (int i = 1; i < n; i++) {
    while (!s.empty() && A[s.top()] <= A[i]) s.pop();
    if (!s.empty())
      l[i] = i - s.top() - 1;
    else
      l[i] = i;
    s.push(i);
  }
  for (int i = k; i <= n; i += (k - 1)) dp[i] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i] += dp[i - 1];
  }
  for (int i = 1; i <= n; i++) dp[i] = (dp[i] + dp[i - 1]) % 1000000007;
  long long int sum = 0;
  long long int temp;
  for (int i = 0; i < n; i++) {
    temp = (dp[l[i] + r[i] + 1] - dp[r[i]] - dp[l[i]]) % 1000000007;
    if (temp < 0) temp += 1000000007;
    sum = (sum + 1LL * temp * A[i]) % 1000000007;
  }
  sum = sum % 1000000007;
  cout << sum << "\n";
}

```