#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
int sum(int a, int b) { return (a + b) % MOD; }
int diff(int a, int b) { return (a + MOD - b) % MOD; }
int prod(int a, int b) { return a * 1ll * b % MOD; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int N, K;
  cin >> N >> K;
  vector<int> vals(N);
  for (int i = 0; i < N; i++) cin >> vals[i];
  priority_queue<pair<int, int>> pq;
  vector<int> maxk(N, 0);
  for (int i = N - 1; i >= 0; i--) {
    pq.push({vals[i], i});
    while (pq.top().second >= i + K) pq.pop();
    maxk[i] = pq.top().first;
  }
  K--;
  int fin = 0;
  vector<int> res(K);
  vector<vector<pair<int, int>>> pf(K);
  for (int i = N - K - 1; i >= 0; i--) {
    int &ans = res[i % K];
    vector<pair<int, int>> &fam = pf[i % K];
    if (fam.empty()) {
      fam.emplace_back(INT_MAX, i + K);
    }
    int F = fam.size();
    while (fam[F - 1].first <= maxk[i]) {
      ans = diff(ans, prod(fam[F - 1].first,
                           (fam[F - 2].second - fam[F - 1].second) / K));
      fam.pop_back();
      F--;
    }
    ans = sum(ans, prod(maxk[i], (fam[F - 1].second - i) / K));
    fam.emplace_back(maxk[i], i);
    fin = sum(fin, ans);
  }
  cout << fin << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
int sum(int a, int b) { return (a + b) % MOD; }
int diff(int a, int b) { return (a + MOD - b) % MOD; }
int prod(int a, int b) { return a * 1ll * b % MOD; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);
  int N, K;
  cin >> N >> K;
  vector<int> vals(N);
  for (int i = 0; i < N; i++) cin >> vals[i];
  priority_queue<pair<int, int>> pq;
  vector<int> maxk(N, 0);
  for (int i = N - 1; i >= 0; i--) {
    pq.push({vals[i], i});
    while (pq.top().second >= i + K) pq.pop();
    maxk[i] = pq.top().first;
  }
  K--;
  int fin = 0;
  vector<int> res(K);
  vector<vector<pair<int, int>>> pf(K);
  for (int i = N - K - 1; i >= 0; i--) {
    int &ans = res[i % K];
    vector<pair<int, int>> &fam = pf[i % K];
    if (fam.empty()) {
      fam.emplace_back(INT_MAX, i + K);
    }
    int F = fam.size();
    while (fam[F - 1].first <= maxk[i]) {
      ans = diff(ans, prod(fam[F - 1].first,
                           (fam[F - 2].second - fam[F - 1].second) / K));
      fam.pop_back();
      F--;
    }
    ans = sum(ans, prod(maxk[i], (fam[F - 1].second - i) / K));
    fam.emplace_back(maxk[i], i);
    fin = sum(fin, ans);
  }
  cout << fin << endl;
  return 0;
}

```