#include <bits/stdc++.h>
using namespace std;
long long read() {
  register long long x = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * w;
}
long long a[500005 << 2], mod, k;
int l[500005 << 2], r[500005 << 2];
long long lgcd(long long a, long long b, long long c, long long n) {
  if (!a) return 0;
  if (a >= c || b >= c) {
    return (((a / c) * (n * (n + 1) / 2) + (b / c) * (n + 1)) % mod +
            lgcd(a % c, b % c, c, n) % mod) %
           mod;
  }
  long long m = (a * n + b) / c;
  return (((m % mod) * (n % mod)) % mod - lgcd(c, c - b - 1, a, m - 1) % mod +
          mod) %
         mod;
}
long long work(int i) {
  return (lgcd(1, r[i] - i, k - 1, i - l[i]) - lgcd(1, 0, k - 1, i - l[i] - 1) +
          mod) %
         mod;
}
int main() {
  long long n = read();
  k = read();
  mod = 1000000007;
  long long ans = 0;
  for (register int i = 1; i <= n; ++i) a[i] = read();
  l[1] = 1, r[n] = n;
  for (int i = 2; i <= n; ++i) {
    int x = i;
    while (x > 1 && a[i] >= a[x - 1]) x = l[x - 1];
    l[i] = x;
  }
  for (int i = n - 1; i >= 1; --i) {
    register int x = i;
    while (x < n && a[i] > a[x + 1]) x = r[x + 1];
    r[i] = x;
  }
  for (register int i = 1; i <= n; ++i) {
    ans = (ans + ((work(i) % mod) * (a[i] % mod)) % mod) % mod;
  }
  cout << ans;
}

### 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;
long long read() {
  register long long x = 0, w = 1;
  register char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 3) + (x << 1) + ch - '0';
    ch = getchar();
  }
  return x * w;
}
long long a[500005 << 2], mod, k;
int l[500005 << 2], r[500005 << 2];
long long lgcd(long long a, long long b, long long c, long long n) {
  if (!a) return 0;
  if (a >= c || b >= c) {
    return (((a / c) * (n * (n + 1) / 2) + (b / c) * (n + 1)) % mod +
            lgcd(a % c, b % c, c, n) % mod) %
           mod;
  }
  long long m = (a * n + b) / c;
  return (((m % mod) * (n % mod)) % mod - lgcd(c, c - b - 1, a, m - 1) % mod +
          mod) %
         mod;
}
long long work(int i) {
  return (lgcd(1, r[i] - i, k - 1, i - l[i]) - lgcd(1, 0, k - 1, i - l[i] - 1) +
          mod) %
         mod;
}
int main() {
  long long n = read();
  k = read();
  mod = 1000000007;
  long long ans = 0;
  for (register int i = 1; i <= n; ++i) a[i] = read();
  l[1] = 1, r[n] = n;
  for (int i = 2; i <= n; ++i) {
    int x = i;
    while (x > 1 && a[i] >= a[x - 1]) x = l[x - 1];
    l[i] = x;
  }
  for (int i = n - 1; i >= 1; --i) {
    register int x = i;
    while (x < n && a[i] > a[x + 1]) x = r[x + 1];
    r[i] = x;
  }
  for (register int i = 1; i <= n; ++i) {
    ans = (ans + ((work(i) % mod) * (a[i] % mod)) % mod) % mod;
  }
  cout << ans;
}

```