#include <bits/stdc++.h>
using namespace std;
long long n, k, a[1000001], l[1000001], r[1000001], ans;
long long S(long long k) { return (k * (k + 1) / 2ll) % 1000000007; }
long long f(long long a, long long b, long long c, long long n) {
  if (!a) return 0;
  if (a >= c || b >= c)
    return ((a / c) * S(n) % 1000000007 + (n + 1) * (b / c) % 1000000007 +
            f(a % c, b % c, c, n)) %
           1000000007;
  long long m = (a * n + b) / c;
  return (m * n % 1000000007 - f(c, c - b - 1, a, m - 1) + 1000000007) %
         1000000007;
}
long long get(long long l, long long r, long long x) {
  return (f(1, r - x, k, x - l) - f(1, 0, k, x - l - 1) + 1000000007) %
         1000000007;
}
int main() {
  scanf("%I64d%I64d", &n, &k);
  for (long long i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  l[1] = 1, r[n] = n;
  for (long long i = 2; i <= n; i++) {
    long long _ = i;
    while (_ > 1 && a[i] >= a[_ - 1]) _ = l[_ - 1];
    l[i] = _;
  }
  for (long long i = n - 1; i >= 1; i--) {
    long long _ = i;
    while (_ < n && a[i] > a[_ + 1]) _ = r[_ + 1];
    r[i] = _;
  }
  k--;
  for (long long i = 1; i <= n; i++)
    ans += get(l[i], r[i], i) * a[i] % 1000000007, ans %= 1000000007;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Create a solution in CPP for 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;
long long n, k, a[1000001], l[1000001], r[1000001], ans;
long long S(long long k) { return (k * (k + 1) / 2ll) % 1000000007; }
long long f(long long a, long long b, long long c, long long n) {
  if (!a) return 0;
  if (a >= c || b >= c)
    return ((a / c) * S(n) % 1000000007 + (n + 1) * (b / c) % 1000000007 +
            f(a % c, b % c, c, n)) %
           1000000007;
  long long m = (a * n + b) / c;
  return (m * n % 1000000007 - f(c, c - b - 1, a, m - 1) + 1000000007) %
         1000000007;
}
long long get(long long l, long long r, long long x) {
  return (f(1, r - x, k, x - l) - f(1, 0, k, x - l - 1) + 1000000007) %
         1000000007;
}
int main() {
  scanf("%I64d%I64d", &n, &k);
  for (long long i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  l[1] = 1, r[n] = n;
  for (long long i = 2; i <= n; i++) {
    long long _ = i;
    while (_ > 1 && a[i] >= a[_ - 1]) _ = l[_ - 1];
    l[i] = _;
  }
  for (long long i = n - 1; i >= 1; i--) {
    long long _ = i;
    while (_ < n && a[i] > a[_ + 1]) _ = r[_ + 1];
    r[i] = _;
  }
  k--;
  for (long long i = 1; i <= n; i++)
    ans += get(l[i], r[i], i) * a[i] % 1000000007, ans %= 1000000007;
  printf("%I64d\n", ans);
  return 0;
}

```