#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
int n, k;
long long a[1000010];
vector<vector<int>> inds(1000010);
long long modvals[1000010];
multiset<long long> curin;
long long myans[1000010];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    inds[i % (k - 1)].push_back(i);
  }
  for (int i = n - 1; i >= 0; i--) {
    curin.insert(a[i]);
    if (i + k < n) {
      curin.erase(curin.find(a[i + k]));
    }
    modvals[i] = *curin.rbegin();
  }
  int val;
  for (int i = 0; i < k - 1; i++) {
    stack<pair<long long, int>> cur;
    for (int j = inds[i].size() - 1; j >= 0; j--) {
      val = inds[i][j];
      long long cv = modvals[val];
      while (cur.size() && cv >= cur.top().first) {
        cur.pop();
      }
      if (cur.size()) {
        int indo = cur.top().second;
        long long ca = ((indo - j) * cv) % mod;
        ca = (ca + myans[inds[i][indo]]);
        myans[val] = ca;
      } else {
        long long nr = inds[i].size() - 1 - j;
        myans[val] = (nr * cv) % mod;
      }
      cur.push(pair<long long, int>(cv, j));
    }
  }
  long long ans = 0LL;
  for (int i = 0; i < n; i++) {
    ans = (ans + myans[i]) % mod;
  }
  cout << ans << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 long long mod = 1000000007;
int n, k;
long long a[1000010];
vector<vector<int>> inds(1000010);
long long modvals[1000010];
multiset<long long> curin;
long long myans[1000010];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n; i++) {
    inds[i % (k - 1)].push_back(i);
  }
  for (int i = n - 1; i >= 0; i--) {
    curin.insert(a[i]);
    if (i + k < n) {
      curin.erase(curin.find(a[i + k]));
    }
    modvals[i] = *curin.rbegin();
  }
  int val;
  for (int i = 0; i < k - 1; i++) {
    stack<pair<long long, int>> cur;
    for (int j = inds[i].size() - 1; j >= 0; j--) {
      val = inds[i][j];
      long long cv = modvals[val];
      while (cur.size() && cv >= cur.top().first) {
        cur.pop();
      }
      if (cur.size()) {
        int indo = cur.top().second;
        long long ca = ((indo - j) * cv) % mod;
        ca = (ca + myans[inds[i][indo]]);
        myans[val] = ca;
      } else {
        long long nr = inds[i].size() - 1 - j;
        myans[val] = (nr * cv) % mod;
      }
      cur.push(pair<long long, int>(cv, j));
    }
  }
  long long ans = 0LL;
  for (int i = 0; i < n; i++) {
    ans = (ans + myans[i]) % mod;
  }
  cout << ans << endl;
}

```