#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
const long long MOD = 1e9 + 7;
int n, k, pr[N], nx[N];
long long ans, cnt[N];
vector<int> q, ms;
void prepare() {
  stack<int> s;
  for (int i = 0; i < n; ++i) {
    while (!s.empty() && ms[s.top()] <= ms[i]) s.pop();
    pr[i] = (s.empty() ? 0 : s.top() + 1);
    s.push(i);
  }
  while (!s.empty()) s.pop();
  for (int i = n - 1; i >= 0; --i) {
    while (!s.empty() && ms[s.top()] < ms[i]) s.pop();
    nx[i] = (s.empty() ? n - 1 : s.top() - 1);
    s.push(i);
  }
  for (int i = 1; i * (k - 1) + 1 <= n; ++i) q.push_back(i * (k - 1) + 1);
  int p = 0;
  long long sum = 0;
  for (int i = 1; i <= n; ++i) {
    while (p < (int)q.size() && q[p] <= i) sum += q[p++];
    cnt[i] = (1ll * i * p - sum + p) % MOD;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    ms.push_back(x);
  }
  prepare();
  for (int i = 0; i < n; ++i) {
    ans += 1ll * ms[i] *
           (cnt[nx[i] - pr[i] + 1] - cnt[i - pr[i]] - cnt[nx[i] - i]) % MOD;
    ans %= MOD;
  }
  printf("%lld", ans);
  return 0;
}

### 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 int N = 1e6 + 5;
const long long MOD = 1e9 + 7;
int n, k, pr[N], nx[N];
long long ans, cnt[N];
vector<int> q, ms;
void prepare() {
  stack<int> s;
  for (int i = 0; i < n; ++i) {
    while (!s.empty() && ms[s.top()] <= ms[i]) s.pop();
    pr[i] = (s.empty() ? 0 : s.top() + 1);
    s.push(i);
  }
  while (!s.empty()) s.pop();
  for (int i = n - 1; i >= 0; --i) {
    while (!s.empty() && ms[s.top()] < ms[i]) s.pop();
    nx[i] = (s.empty() ? n - 1 : s.top() - 1);
    s.push(i);
  }
  for (int i = 1; i * (k - 1) + 1 <= n; ++i) q.push_back(i * (k - 1) + 1);
  int p = 0;
  long long sum = 0;
  for (int i = 1; i <= n; ++i) {
    while (p < (int)q.size() && q[p] <= i) sum += q[p++];
    cnt[i] = (1ll * i * p - sum + p) % MOD;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    ms.push_back(x);
  }
  prepare();
  for (int i = 0; i < n; ++i) {
    ans += 1ll * ms[i] *
           (cnt[nx[i] - pr[i] + 1] - cnt[i - pr[i]] - cnt[nx[i] - i]) % MOD;
    ans %= MOD;
  }
  printf("%lld", ans);
  return 0;
}

```