#include <bits/stdc++.h>
using namespace std;
const int max_n = 1000111, max_lev = 20, mod = 1000000007;
int n, k, a[max_n], mx[max_lev][max_n], dp[max_n];
multiset<int> q;
void add(int &x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = n - k + 1; i < n; ++i) {
    q.insert(a[i]);
  }
  for (int i = n - k; i >= 0; --i) {
    q.insert(a[i]);
    mx[0][i] = *q.rbegin();
    for (int lev = 1; lev < max_lev; ++lev) {
      long long nxt = i + (1LL << (lev - 1)) * k;
      int val = mx[lev - 1][i];
      if (nxt < n) {
        val = max(val, mx[lev - 1][nxt]);
      }
      mx[lev][i] = val;
    }
    int cur = i;
    for (int lev = max_lev - 1; lev >= 0; --lev) {
      long long nxt = cur + (1LL << lev) * k;
      if (nxt <= n && mx[lev][cur] <= mx[0][i]) {
        cur = nxt;
      }
    }
    dp[i] = (1LL * mx[0][i] * ((cur - i) / k)) % mod;
    if (cur < n) {
      add(dp[i], dp[cur]);
    }
    q.erase(q.find(a[i + k - 1]));
  }
  int ans = 0;
  for (int i = 0; i + k < n; ++i) {
    int first = max(a[i], mx[0][i + 1]);
    int cur = i + k + 1;
    for (int lev = max_lev - 1; lev >= 0; --lev) {
      long long nxt = cur + (1LL << lev) * k;
      if (nxt <= n && mx[lev][cur] <= first) {
        cur = nxt;
      }
    }
    int pl = (1LL * first * (1 + (cur - i - k - 1) / k)) % mod;
    if (cur < n) {
      add(pl, dp[cur]);
    }
    add(ans, pl);
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 max_n = 1000111, max_lev = 20, mod = 1000000007;
int n, k, a[max_n], mx[max_lev][max_n], dp[max_n];
multiset<int> q;
void add(int &x, int y) {
  x += y;
  if (x >= mod) {
    x -= mod;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  --k;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = n - k + 1; i < n; ++i) {
    q.insert(a[i]);
  }
  for (int i = n - k; i >= 0; --i) {
    q.insert(a[i]);
    mx[0][i] = *q.rbegin();
    for (int lev = 1; lev < max_lev; ++lev) {
      long long nxt = i + (1LL << (lev - 1)) * k;
      int val = mx[lev - 1][i];
      if (nxt < n) {
        val = max(val, mx[lev - 1][nxt]);
      }
      mx[lev][i] = val;
    }
    int cur = i;
    for (int lev = max_lev - 1; lev >= 0; --lev) {
      long long nxt = cur + (1LL << lev) * k;
      if (nxt <= n && mx[lev][cur] <= mx[0][i]) {
        cur = nxt;
      }
    }
    dp[i] = (1LL * mx[0][i] * ((cur - i) / k)) % mod;
    if (cur < n) {
      add(dp[i], dp[cur]);
    }
    q.erase(q.find(a[i + k - 1]));
  }
  int ans = 0;
  for (int i = 0; i + k < n; ++i) {
    int first = max(a[i], mx[0][i + 1]);
    int cur = i + k + 1;
    for (int lev = max_lev - 1; lev >= 0; --lev) {
      long long nxt = cur + (1LL << lev) * k;
      if (nxt <= n && mx[lev][cur] <= first) {
        cur = nxt;
      }
    }
    int pl = (1LL * first * (1 + (cur - i - k - 1) / k)) % mod;
    if (cur < n) {
      add(pl, dp[cur]);
    }
    add(ans, pl);
  }
  printf("%d\n", ans);
  return 0;
}

```