#include <bits/stdc++.h>
using namespace std;
const int INF = 1000 * 1000 * 1000;
long long n, m, k, x, y;
pair<long long, long long> a[1005000];
long long nx[1005000], pr[1005000];
long long cnt(long long l, long long r) {
  if (l > r) swap(l, r);
  long long sum = l / (k - 1);
  long long ans = 0;
  long long st = 1;
  ans += (st + (st + sum * (k - 1))) * (sum + 1) / 2;
  l -= sum * (k - 1);
  r -= sum * (k - 1);
  st += sum * (k - 1);
  long long dots_l = l;
  long long dots_r = min(r, k - 1);
  st -= (k - 1 - dots_l);
  st -= (k - 1 - dots_r);
  st += k - 1;
  if (st > 0)
    ans += st;
  else
    return ans;
  r -= dots_r;
  sum = r / (k - 1);
  ans += sum * st;
  r -= sum * (k - 1);
  st -= (k - 1 - r);
  if (st > 0)
    ans += st;
  else
    return ans;
  st -= (k - 1);
  if (st <= 0) return ans;
  sum = st / (k - 1);
  ans += (st + (st - sum * (k - 1))) * (sum + 1) / 2;
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i].first = x;
    a[i].second = i;
    nx[i] = i + 1;
    pr[i] = i - 1;
  }
  sort(a + 1, a + n + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long val = a[i].first;
    long long pos = a[i].second;
    long long np = nx[pos];
    long long pp = pr[pos];
    long long l = pos - pp - 1;
    long long r = np - pos - 1;
    ans += (cnt(l, r) - 1) % 1000000007 * val % 1000000007;
    nx[pr[pos]] = np;
    pr[nx[pos]] = pp;
  }
  cout << ans % 1000000007 << endl;
  return 0;
}

### 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 INF = 1000 * 1000 * 1000;
long long n, m, k, x, y;
pair<long long, long long> a[1005000];
long long nx[1005000], pr[1005000];
long long cnt(long long l, long long r) {
  if (l > r) swap(l, r);
  long long sum = l / (k - 1);
  long long ans = 0;
  long long st = 1;
  ans += (st + (st + sum * (k - 1))) * (sum + 1) / 2;
  l -= sum * (k - 1);
  r -= sum * (k - 1);
  st += sum * (k - 1);
  long long dots_l = l;
  long long dots_r = min(r, k - 1);
  st -= (k - 1 - dots_l);
  st -= (k - 1 - dots_r);
  st += k - 1;
  if (st > 0)
    ans += st;
  else
    return ans;
  r -= dots_r;
  sum = r / (k - 1);
  ans += sum * st;
  r -= sum * (k - 1);
  st -= (k - 1 - r);
  if (st > 0)
    ans += st;
  else
    return ans;
  st -= (k - 1);
  if (st <= 0) return ans;
  sum = st / (k - 1);
  ans += (st + (st - sum * (k - 1))) * (sum + 1) / 2;
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i].first = x;
    a[i].second = i;
    nx[i] = i + 1;
    pr[i] = i - 1;
  }
  sort(a + 1, a + n + 1);
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    long long val = a[i].first;
    long long pos = a[i].second;
    long long np = nx[pos];
    long long pp = pr[pos];
    long long l = pos - pp - 1;
    long long r = np - pos - 1;
    ans += (cnt(l, r) - 1) % 1000000007 * val % 1000000007;
    nx[pr[pos]] = np;
    pr[nx[pos]] = pp;
  }
  cout << ans % 1000000007 << endl;
  return 0;
}

```