#include <bits/stdc++.h>
using namespace std;
int n, k;
int a[1000010], re[1000010], pre[1000010], nxt[1000010];
int lsh[1000010];
int t[1000010];
long long ans, s[1000010];
int find(int x) {
  int sum = 0;
  for (; x; x -= (x & -x)) sum = max(sum, t[x]);
  return sum;
}
void add(int x, int k) {
  for (; x <= n; x += (x & -x)) t[x] = max(t[x], k);
}
int main() {
  int L, R, mid;
  scanf("%d%d", &n, &k);
  s[k] = 1;
  for (int i = k + 1; i <= n; i++)
    s[i] = (s[i - 1] + 1 + (i - k) / (k - 1)) % 1000000007;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    lsh[i] = a[i];
  }
  sort(lsh + 1, lsh + 1 + n);
  for (int i = 1; i <= n; i++) {
    L = 1;
    R = n;
    while (L < R) {
      mid = (L + R) >> 1;
      if (lsh[mid] >= a[i])
        R = mid;
      else
        L = mid + 1;
    }
    re[i] = a[i];
    a[i] = L;
  }
  for (int i = 1; i <= n; i++) {
    pre[i] = find(n - a[i] + 1);
    add(n - a[i] + 1, i);
  }
  for (int i = 1; i <= n; i++) t[i] = 0;
  for (int i = n; i; i--) {
    nxt[i] = n - find(n - a[i]) + 1;
    add(n - a[i] + 1, n - i + 1);
  }
  for (int i = 1; i <= n; i++)
    (ans += (long long)re[i] *
            (s[nxt[i] - pre[i] - 1] - s[i - pre[i] - 1] - s[nxt[i] - i - 1]) %
            1000000007) %= 1000000007;
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
int a[1000010], re[1000010], pre[1000010], nxt[1000010];
int lsh[1000010];
int t[1000010];
long long ans, s[1000010];
int find(int x) {
  int sum = 0;
  for (; x; x -= (x & -x)) sum = max(sum, t[x]);
  return sum;
}
void add(int x, int k) {
  for (; x <= n; x += (x & -x)) t[x] = max(t[x], k);
}
int main() {
  int L, R, mid;
  scanf("%d%d", &n, &k);
  s[k] = 1;
  for (int i = k + 1; i <= n; i++)
    s[i] = (s[i - 1] + 1 + (i - k) / (k - 1)) % 1000000007;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    lsh[i] = a[i];
  }
  sort(lsh + 1, lsh + 1 + n);
  for (int i = 1; i <= n; i++) {
    L = 1;
    R = n;
    while (L < R) {
      mid = (L + R) >> 1;
      if (lsh[mid] >= a[i])
        R = mid;
      else
        L = mid + 1;
    }
    re[i] = a[i];
    a[i] = L;
  }
  for (int i = 1; i <= n; i++) {
    pre[i] = find(n - a[i] + 1);
    add(n - a[i] + 1, i);
  }
  for (int i = 1; i <= n; i++) t[i] = 0;
  for (int i = n; i; i--) {
    nxt[i] = n - find(n - a[i]) + 1;
    add(n - a[i] + 1, n - i + 1);
  }
  for (int i = 1; i <= n; i++)
    (ans += (long long)re[i] *
            (s[nxt[i] - pre[i] - 1] - s[i - pre[i] - 1] - s[nxt[i] - i - 1]) %
            1000000007) %= 1000000007;
  printf("%lld\n", ans);
  return 0;
}

```