#include <bits/stdc++.h>
using namespace std;
int dp[1001000], n, K, ans, sum0;
pair<int, int> a[1001000];
set<int> st;
int cal(int len) { return dp[len]; }
void ins(int x) {
  auto it = st.insert(x).first;
  int l = it == st.begin() ? 0 : *prev(it);
  int r = next(it) == st.end() ? n + 1 : *next(it);
  sum0 = (sum0 - cal(r - l - 1) + 1000000007) % 1000000007;
  sum0 = (sum0 + cal(x - l - 1)) % 1000000007;
  sum0 = (sum0 + cal(r - x - 1)) % 1000000007;
}
int main() {
  scanf("%d%d", &n, &K);
  for (int i = K; i <= n; ++i)
    dp[i] = ((i - K + 1) + dp[i - K + 1]) % 1000000007;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].first), a[i].second = i;
  sort(a + 1, a + n + 1);
  for (int i = n, j; i >= 1; i = j) {
    int sum1 = sum0;
    for (j = i; a[j].first == a[i].first && j >= 1; j--) ins(a[j].second);
    ans = (ans + 1ll * (sum1 - sum0 + 1000000007) * a[i].first) % 1000000007;
  }
  printf("%d", ans);
}

### 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;
int dp[1001000], n, K, ans, sum0;
pair<int, int> a[1001000];
set<int> st;
int cal(int len) { return dp[len]; }
void ins(int x) {
  auto it = st.insert(x).first;
  int l = it == st.begin() ? 0 : *prev(it);
  int r = next(it) == st.end() ? n + 1 : *next(it);
  sum0 = (sum0 - cal(r - l - 1) + 1000000007) % 1000000007;
  sum0 = (sum0 + cal(x - l - 1)) % 1000000007;
  sum0 = (sum0 + cal(r - x - 1)) % 1000000007;
}
int main() {
  scanf("%d%d", &n, &K);
  for (int i = K; i <= n; ++i)
    dp[i] = ((i - K + 1) + dp[i - K + 1]) % 1000000007;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i].first), a[i].second = i;
  sort(a + 1, a + n + 1);
  for (int i = n, j; i >= 1; i = j) {
    int sum1 = sum0;
    for (j = i; a[j].first == a[i].first && j >= 1; j--) ins(a[j].second);
    ans = (ans + 1ll * (sum1 - sum0 + 1000000007) * a[i].first) % 1000000007;
  }
  printf("%d", ans);
}

```