#include <bits/stdc++.h>
using namespace std;
const int N = 1e6;
int n, k;
vector<pair<long long int, long long int> > q;
long long int ans = 0;
const long long int MOD = 1e9 + 7;
long long int func(long long int l, long long int x) {
  long long int c = (l - 1) / (k - 1);
  long long int d = (l - x - 1) / (k - 1) + 1;
  if (l == x) d = 1;
  long long int e = (x - 1 - 1) / (k - 1) + 1;
  if (x == 1) e = 1;
  long long int ans1 = ((d - 1) * x) % MOD + (l * (c - d + 1)) % MOD -
                       (((c + d) * (c - d + 1) / 2) % MOD * (k - 1)) % MOD;
  ans1 = (ans1 + MOD) % MOD;
  long long int ans2 = (x * (e - 1)) % MOD -
                       ((e * (e - 1) / 2) % MOD * (k - 1)) % MOD + (c - e + 1);
  ans2 = (ans2 + MOD) % MOD;
  long long int ans = (ans1 + MOD - ans2 + c) % MOD;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long int tmp;
    cin >> tmp;
    while (!q.empty() && q[q.size() - 1].first <= tmp) {
      if (q.size() == 1) {
        ans = (ans +
               (q[q.size() - 1].first * func(i, q[q.size() - 1].second + 1)) %
                   MOD) %
              MOD;
      } else {
        ans = (ans + (q[q.size() - 1].first *
                      func(i - q[q.size() - 2].second - 1,
                           q[q.size() - 1].second - q[q.size() - 2].second)) %
                         MOD) %
              MOD;
      }
      q.pop_back();
    }
    q.push_back(make_pair(tmp, i));
  }
  while (!q.empty()) {
    if (q.size() == 1) {
      ans =
          (ans + (q[q.size() - 1].first * func(n, q[q.size() - 1].second + 1)) %
                     MOD) %
          MOD;
    } else {
      ans = (ans + (q[q.size() - 1].first *
                    func(n - q[q.size() - 2].second - 1,
                         q[q.size() - 1].second - q[q.size() - 2].second)) %
                       MOD) %
            MOD;
    }
    q.pop_back();
  }
  cout << ans << endl;
}

### 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;
const int N = 1e6;
int n, k;
vector<pair<long long int, long long int> > q;
long long int ans = 0;
const long long int MOD = 1e9 + 7;
long long int func(long long int l, long long int x) {
  long long int c = (l - 1) / (k - 1);
  long long int d = (l - x - 1) / (k - 1) + 1;
  if (l == x) d = 1;
  long long int e = (x - 1 - 1) / (k - 1) + 1;
  if (x == 1) e = 1;
  long long int ans1 = ((d - 1) * x) % MOD + (l * (c - d + 1)) % MOD -
                       (((c + d) * (c - d + 1) / 2) % MOD * (k - 1)) % MOD;
  ans1 = (ans1 + MOD) % MOD;
  long long int ans2 = (x * (e - 1)) % MOD -
                       ((e * (e - 1) / 2) % MOD * (k - 1)) % MOD + (c - e + 1);
  ans2 = (ans2 + MOD) % MOD;
  long long int ans = (ans1 + MOD - ans2 + c) % MOD;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long int tmp;
    cin >> tmp;
    while (!q.empty() && q[q.size() - 1].first <= tmp) {
      if (q.size() == 1) {
        ans = (ans +
               (q[q.size() - 1].first * func(i, q[q.size() - 1].second + 1)) %
                   MOD) %
              MOD;
      } else {
        ans = (ans + (q[q.size() - 1].first *
                      func(i - q[q.size() - 2].second - 1,
                           q[q.size() - 1].second - q[q.size() - 2].second)) %
                         MOD) %
              MOD;
      }
      q.pop_back();
    }
    q.push_back(make_pair(tmp, i));
  }
  while (!q.empty()) {
    if (q.size() == 1) {
      ans =
          (ans + (q[q.size() - 1].first * func(n, q[q.size() - 1].second + 1)) %
                     MOD) %
          MOD;
    } else {
      ans = (ans + (q[q.size() - 1].first *
                    func(n - q[q.size() - 2].second - 1,
                         q[q.size() - 1].second - q[q.size() - 2].second)) %
                       MOD) %
            MOD;
    }
    q.pop_back();
  }
  cout << ans << endl;
}

```