#include <bits/stdc++.h>
using namespace std;
inline int read() {
  register int data = 0, w = 1;
  char ch = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9')
    data = (data << 1) + (data << 3) + (ch ^ 48), ch = getchar();
  return data * w;
}
int n, k, b[2000005], a[2000005];
long long ans, s[2000005], fir[2000005];
const int mod = 1e9 + 7;
void dicon(int l, int r) {
  if (r - l < k - 1) return;
  int mid = l + r >> 1;
  b[mid] = a[mid];
  for (int i = mid - 1; i >= l; i--) b[i] = max(a[i], b[i + 1]);
  b[mid + 1] = a[mid + 1];
  for (int i = mid + 2; i <= r; i++) b[i] = max(b[i - 1], a[i]);
  for (int i = r; i >= mid + 1; i--)
    s[i] = (b[i] + s[i + k - 1]) % mod, fir[i % (k - 1)] = i;
  for (int i = mid; i >= l && i + k - 1 > mid; i--) {
    int t = fir[i % (k - 1)];
    if (!t) continue;
    for (int j = i; j >= l; j -= k - 1) {
      while (t <= r && b[t] <= b[j]) t += k - 1;
      ans = (ans + s[t] + 1ll * (t - fir[i % (k - 1)]) / (k - 1) * b[j]) % mod;
    }
  }
  for (int i = mid + 1; i <= r; i++) fir[i % (k - 1)] = s[i] = 0;
  dicon(l, mid);
  dicon(mid + 1, r);
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  dicon(1, n);
  cout << ans << '\n';
  return 0;
}

### Prompt

Your task is to create 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;
inline int read() {
  register int data = 0, w = 1;
  char ch = 0;
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9')
    data = (data << 1) + (data << 3) + (ch ^ 48), ch = getchar();
  return data * w;
}
int n, k, b[2000005], a[2000005];
long long ans, s[2000005], fir[2000005];
const int mod = 1e9 + 7;
void dicon(int l, int r) {
  if (r - l < k - 1) return;
  int mid = l + r >> 1;
  b[mid] = a[mid];
  for (int i = mid - 1; i >= l; i--) b[i] = max(a[i], b[i + 1]);
  b[mid + 1] = a[mid + 1];
  for (int i = mid + 2; i <= r; i++) b[i] = max(b[i - 1], a[i]);
  for (int i = r; i >= mid + 1; i--)
    s[i] = (b[i] + s[i + k - 1]) % mod, fir[i % (k - 1)] = i;
  for (int i = mid; i >= l && i + k - 1 > mid; i--) {
    int t = fir[i % (k - 1)];
    if (!t) continue;
    for (int j = i; j >= l; j -= k - 1) {
      while (t <= r && b[t] <= b[j]) t += k - 1;
      ans = (ans + s[t] + 1ll * (t - fir[i % (k - 1)]) / (k - 1) * b[j]) % mod;
    }
  }
  for (int i = mid + 1; i <= r; i++) fir[i % (k - 1)] = s[i] = 0;
  dicon(l, mid);
  dicon(mid + 1, r);
}
int main() {
  n = read(), k = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  dicon(1, n);
  cout << ans << '\n';
  return 0;
}

```