#include <bits/stdc++.h>
using namespace std;
const int N = (int)1e6 + 7;
const int MOD = (int)1e9 + 7;
const long long INF = (long long)1e18 + 7;
long long a[N];
int L[N], R[N];
int n, k;
long long calc(int l, int r) {
  long long n = r - l + 1;
  if (n < k) return 0;
  long long x = (n - 1) / (k - 1);
  return (n * x - (k - 1LL) * x * (x + 1LL) / 2LL) % MOD;
  long long ans = x * (x + 1LL) / 2;
  ans = n * x - (k - 1) * ans;
  return ans % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  deque<int> q;
  a[0] = a[n + 1] = MOD;
  q.push_back(0);
  for (int i = 1; i <= n; i++) {
    while (a[q.back()] < a[i]) q.pop_back();
    L[i] = q.back() + 1;
    q.push_back(i);
  }
  q = {n + 1};
  for (int i = n; i >= 1; i--) {
    while (a[q.back()] <= a[i]) q.pop_back();
    R[i] = q.back() - 1;
    q.push_back(i);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long num = calc(L[i], R[i]) - calc(L[i], i - 1) - calc(i + 1, R[i]);
    ans += a[i] * num;
    ans %= MOD;
  }
  cout << ans;
}

### Prompt

Develop a solution in CPP 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 int N = (int)1e6 + 7;
const int MOD = (int)1e9 + 7;
const long long INF = (long long)1e18 + 7;
long long a[N];
int L[N], R[N];
int n, k;
long long calc(int l, int r) {
  long long n = r - l + 1;
  if (n < k) return 0;
  long long x = (n - 1) / (k - 1);
  return (n * x - (k - 1LL) * x * (x + 1LL) / 2LL) % MOD;
  long long ans = x * (x + 1LL) / 2;
  ans = n * x - (k - 1) * ans;
  return ans % MOD;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  deque<int> q;
  a[0] = a[n + 1] = MOD;
  q.push_back(0);
  for (int i = 1; i <= n; i++) {
    while (a[q.back()] < a[i]) q.pop_back();
    L[i] = q.back() + 1;
    q.push_back(i);
  }
  q = {n + 1};
  for (int i = n; i >= 1; i--) {
    while (a[q.back()] <= a[i]) q.pop_back();
    R[i] = q.back() - 1;
    q.push_back(i);
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long num = calc(L[i], R[i]) - calc(L[i], i - 1) - calc(i + 1, R[i]);
    ans += a[i] * num;
    ans %= MOD;
  }
  cout << ans;
}

```