#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 5;
int n, k;
int a[maxn];
int lp[maxn], rp[maxn];
int main() {
  ios::sync_with_stdio(false);
  while (cin >> n >> k) {
    for (int i = 1; i <= n; i++) cin >> a[i];
    deque<int> q;
    for (int i = 1; i <= n; i++) {
      while (!q.empty() && a[q.back()] < a[i]) q.pop_back();
      lp[i] = q.empty() ? 1 : q.back() + 1;
      q.push_back(i);
    }
    deque<int> q2;
    for (int i = n; i >= 1; i--) {
      while (!q2.empty() && a[q2.back()] <= a[i]) q2.pop_back();
      rp[i] = q2.empty() ? n : q2.back() - 1;
      q2.push_back(i);
    }
    auto cal = [&](int l, int r) -> long long {
      if (r - l + 1 < k) return 0;
      long long len = r - l + 1;
      long long k1 = len - k + 1;
      long long add = k1 / (k - 1);
      long long k2 = k1 - add * (k - 1);
      long long ans = (k1 + k2) * (1 + add) / 2;
      return ans % mod;
    };
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      long long res =
          (cal(lp[i], rp[i]) - cal(lp[i], i - 1) - cal(i + 1, rp[i])) % mod;
      ans = (ans + res * a[i]) % mod;
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Generate 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 mod = 1e9 + 7;
const int maxn = 1e6 + 5;
int n, k;
int a[maxn];
int lp[maxn], rp[maxn];
int main() {
  ios::sync_with_stdio(false);
  while (cin >> n >> k) {
    for (int i = 1; i <= n; i++) cin >> a[i];
    deque<int> q;
    for (int i = 1; i <= n; i++) {
      while (!q.empty() && a[q.back()] < a[i]) q.pop_back();
      lp[i] = q.empty() ? 1 : q.back() + 1;
      q.push_back(i);
    }
    deque<int> q2;
    for (int i = n; i >= 1; i--) {
      while (!q2.empty() && a[q2.back()] <= a[i]) q2.pop_back();
      rp[i] = q2.empty() ? n : q2.back() - 1;
      q2.push_back(i);
    }
    auto cal = [&](int l, int r) -> long long {
      if (r - l + 1 < k) return 0;
      long long len = r - l + 1;
      long long k1 = len - k + 1;
      long long add = k1 / (k - 1);
      long long k2 = k1 - add * (k - 1);
      long long ans = (k1 + k2) * (1 + add) / 2;
      return ans % mod;
    };
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      long long res =
          (cal(lp[i], rp[i]) - cal(lp[i], i - 1) - cal(i + 1, rp[i])) % mod;
      ans = (ans + res * a[i]) % mod;
    }
    cout << ans << endl;
  }
  return 0;
}

```