#include <bits/stdc++.h>
using namespace std;
const long long sz = 1e6 + 4;
long long a[sz], ind[sz], sums[sz];
stack<long long> s;
long long n, k;
long long sms(long long l, long long r) {
  if (l > r) return 0;
  long long cff = (r - l + 1) / k;
  return cff * (r - l + 1) - (cff * (cff + 1) / 2) * k;
}
void solve() {
  cin >> n >> k;
  --k;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = n - 1; i >= 0; --i) {
    while (s.size() && a[s.top()] <= a[i]) {
      s.pop();
    }
    if (s.size()) {
      ind[i] = s.top();
    } else {
      ind[i] = n;
    }
    s.push(i);
  }
  while (s.size()) {
    s.pop();
  }
  long long l = -1;
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    while (s.size() && a[s.top()] < a[i]) {
      s.pop();
    }
    if (s.size()) {
      l = s.top() + 1;
    } else {
      l = 0;
    }
    long long num = 0;
    long long r = ind[i] - 1;
    num = sms(l, r) - sms(i + 1, r) - sms(l, i - 1);
    ans += (num % 1000000007) * (a[i]);
    ans %= 1000000007;
    s.push(i);
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const long long sz = 1e6 + 4;
long long a[sz], ind[sz], sums[sz];
stack<long long> s;
long long n, k;
long long sms(long long l, long long r) {
  if (l > r) return 0;
  long long cff = (r - l + 1) / k;
  return cff * (r - l + 1) - (cff * (cff + 1) / 2) * k;
}
void solve() {
  cin >> n >> k;
  --k;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = n - 1; i >= 0; --i) {
    while (s.size() && a[s.top()] <= a[i]) {
      s.pop();
    }
    if (s.size()) {
      ind[i] = s.top();
    } else {
      ind[i] = n;
    }
    s.push(i);
  }
  while (s.size()) {
    s.pop();
  }
  long long l = -1;
  long long ans = 0;
  for (long long i = 0; i < n; ++i) {
    while (s.size() && a[s.top()] < a[i]) {
      s.pop();
    }
    if (s.size()) {
      l = s.top() + 1;
    } else {
      l = 0;
    }
    long long num = 0;
    long long r = ind[i] - 1;
    num = sms(l, r) - sms(i + 1, r) - sms(l, i - 1);
    ans += (num % 1000000007) * (a[i]);
    ans %= 1000000007;
    s.push(i);
  }
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```