#include <bits/stdc++.h>
using namespace std;
int n, k, a[1000010], b[1000010];
bool cmp(int i, int j) { return a[i] < a[j]; }
int prv[1000010], nxt[1000010];
const int mod = 1e9 + 7;
int func(int x) { return max(0, x / (k - 1)); }
int calc(int l, int m, int r) {
  int ans = 0;
  if (m - l > r - m) m = l + r - m;
  int d = func(m - l), u = func(r - m);
  ans = (ans + 1ll * d + 1ll * d * (d + 1) / 2 % mod * (k - 1)) % mod;
  ans = (ans + 1ll * (m - l + 1) * (u - d)) % mod;
  int w = (r - l + 1) / (k - 1);
  ans = (ans + 1ll * (r - l + 1) * (w - u) -
         1ll * (w - u) * (w + u + 1) / 2 % mod * (k - 1)) %
        mod;
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = i;
  }
  sort(b + 1, b + n + 1, cmp);
  for (int i = 1; i <= n; i++) prv[i] = i - 1, nxt[i] = i + 1;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int j = b[i];
    ans = (ans + 1ll * calc(prv[j] + 1, j, nxt[j] - 1) * a[j]) % mod;
    int l = prv[j], r = nxt[j];
    prv[r] = l, nxt[l] = r;
  }
  if (ans < 0) ans += mod;
  cout << ans << endl;
  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;
int n, k, a[1000010], b[1000010];
bool cmp(int i, int j) { return a[i] < a[j]; }
int prv[1000010], nxt[1000010];
const int mod = 1e9 + 7;
int func(int x) { return max(0, x / (k - 1)); }
int calc(int l, int m, int r) {
  int ans = 0;
  if (m - l > r - m) m = l + r - m;
  int d = func(m - l), u = func(r - m);
  ans = (ans + 1ll * d + 1ll * d * (d + 1) / 2 % mod * (k - 1)) % mod;
  ans = (ans + 1ll * (m - l + 1) * (u - d)) % mod;
  int w = (r - l + 1) / (k - 1);
  ans = (ans + 1ll * (r - l + 1) * (w - u) -
         1ll * (w - u) * (w + u + 1) / 2 % mod * (k - 1)) %
        mod;
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    b[i] = i;
  }
  sort(b + 1, b + n + 1, cmp);
  for (int i = 1; i <= n; i++) prv[i] = i - 1, nxt[i] = i + 1;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    int j = b[i];
    ans = (ans + 1ll * calc(prv[j] + 1, j, nxt[j] - 1) * a[j]) % mod;
    int l = prv[j], r = nxt[j];
    prv[r] = l, nxt[l] = r;
  }
  if (ans < 0) ans += mod;
  cout << ans << endl;
  return 0;
}

```