#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int MN = 1000010;
int N, K;
int A[MN];
vector<pair<int, int> > ord;
set<int> st;
int L[MN], R[MN];
int main() {
  scanf("%d %d", &N, &K);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    ord.push_back(pair<int, int>(A[i], i));
  }
  sort(ord.begin(), ord.end());
  for (int i = 0; i < N; i++) {
    L[i] = -1;
    R[i] = N;
  }
  for (int i = (int)ord.size() - 1; i >= 0; i--) {
    int x = ord[i].second;
    auto it = st.upper_bound(x);
    if (it != st.end()) R[x] = *it;
    if (it != st.begin()) L[x] = *(--it);
    st.insert(x);
  }
  int ans = 0;
  for (int i = 0; i < N; i++) {
    int l = i - L[i];
    int r = R[i] - i;
    int len = l + r - 1;
    if (l > r) swap(l, r);
    int tmp = 0;
    int x1 = (l - 1) / (K - 1);
    int x2 = (r - 1) / (K - 1) + 1;
    int x3 = (len - 1) / (K - 1);
    tmp += 1LL * x1 * (x1 + 1) / 2 % mod * K % mod;
    tmp %= mod;
    tmp += mod - 1LL * x1 * (x1 - 1) / 2 % mod;
    tmp %= mod;
    tmp += 1LL * (x2 - x1 - 1) * l % mod;
    tmp %= mod;
    if (x2 <= x3) {
      tmp += 1LL * (len + 1) * (x3 - x2 + 1) % mod;
      tmp %= mod;
      tmp += mod - (x3 - x2 + 1);
      tmp %= mod;
      tmp += 1LL * (x2 + x3) * (x3 - x2 + 1) / 2 % mod;
      tmp %= mod;
      tmp += mod - 1LL * (x2 + x3) * (x3 - x2 + 1) / 2 % mod * K % mod;
      tmp %= mod;
    }
    ans += 1LL * tmp * A[i] % mod;
    ans %= mod;
  }
  printf("%d", ans);
}

### 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;
const int MN = 1000010;
int N, K;
int A[MN];
vector<pair<int, int> > ord;
set<int> st;
int L[MN], R[MN];
int main() {
  scanf("%d %d", &N, &K);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
    ord.push_back(pair<int, int>(A[i], i));
  }
  sort(ord.begin(), ord.end());
  for (int i = 0; i < N; i++) {
    L[i] = -1;
    R[i] = N;
  }
  for (int i = (int)ord.size() - 1; i >= 0; i--) {
    int x = ord[i].second;
    auto it = st.upper_bound(x);
    if (it != st.end()) R[x] = *it;
    if (it != st.begin()) L[x] = *(--it);
    st.insert(x);
  }
  int ans = 0;
  for (int i = 0; i < N; i++) {
    int l = i - L[i];
    int r = R[i] - i;
    int len = l + r - 1;
    if (l > r) swap(l, r);
    int tmp = 0;
    int x1 = (l - 1) / (K - 1);
    int x2 = (r - 1) / (K - 1) + 1;
    int x3 = (len - 1) / (K - 1);
    tmp += 1LL * x1 * (x1 + 1) / 2 % mod * K % mod;
    tmp %= mod;
    tmp += mod - 1LL * x1 * (x1 - 1) / 2 % mod;
    tmp %= mod;
    tmp += 1LL * (x2 - x1 - 1) * l % mod;
    tmp %= mod;
    if (x2 <= x3) {
      tmp += 1LL * (len + 1) * (x3 - x2 + 1) % mod;
      tmp %= mod;
      tmp += mod - (x3 - x2 + 1);
      tmp %= mod;
      tmp += 1LL * (x2 + x3) * (x3 - x2 + 1) / 2 % mod;
      tmp %= mod;
      tmp += mod - 1LL * (x2 + x3) * (x3 - x2 + 1) / 2 % mod * K % mod;
      tmp %= mod;
    }
    ans += 1LL * tmp * A[i] % mod;
    ans %= mod;
  }
  printf("%d", ans);
}

```