#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
const long long mod = 1000000007;
int a[N], sta[N], top;
int l[N], r[N];
void add(long long &x, long long y) {
  x += y;
  while (x >= mod) x -= mod;
}
int neq(int a, int K, int ff) {
  if (ff == 1) return (a - 1) / (K - 1) + 1;
}
long long cal0(long long f, long long n, long long d) {
  if (n <= 0) return 0ll;
  long long g = (f + (n - 1) * d) % mod;
  return ((f + g) * n / 2) % mod;
}
long long cal(int l, int r, int i, int K) {
  if (i - l > r - i) i = r - (i - l);
  long long tm = neq(r - l + 1, K, 1);
  long long tl = neq(i - l + 1, K, 1), tr = neq(r - i + 1, K, 1);
  long long ll = i - l + 1, rr = r - i + 1, mm = r - l + 1;
  long long ans1 = cal0(K, tl - 1, K - 1);
  long long ans2 = ll * (tr - tl) % mod;
  long long f = (mm - (tr * K - tr + 1) + 1) % mod;
  long long n = tm - tr;
  long long ans3 = cal0(f, n, mod - (K - 1));
  return (ans1 + ans2 + ans3) % mod;
}
int main() {
  int i, n, K;
  cin >> n >> K;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    while (top && a[i] > a[sta[top]]) r[sta[top]] = i - 1, top--;
    l[i] = sta[top] + 1, sta[++top] = i;
  }
  for (i = 1; i <= top; i++) r[sta[i]] = n;
  long long ans = 0;
  for (i = 1; i <= n; i++) add(ans, cal(l[i], r[i], i, K) * a[i] % mod);
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const int N = 1000010;
const long long mod = 1000000007;
int a[N], sta[N], top;
int l[N], r[N];
void add(long long &x, long long y) {
  x += y;
  while (x >= mod) x -= mod;
}
int neq(int a, int K, int ff) {
  if (ff == 1) return (a - 1) / (K - 1) + 1;
}
long long cal0(long long f, long long n, long long d) {
  if (n <= 0) return 0ll;
  long long g = (f + (n - 1) * d) % mod;
  return ((f + g) * n / 2) % mod;
}
long long cal(int l, int r, int i, int K) {
  if (i - l > r - i) i = r - (i - l);
  long long tm = neq(r - l + 1, K, 1);
  long long tl = neq(i - l + 1, K, 1), tr = neq(r - i + 1, K, 1);
  long long ll = i - l + 1, rr = r - i + 1, mm = r - l + 1;
  long long ans1 = cal0(K, tl - 1, K - 1);
  long long ans2 = ll * (tr - tl) % mod;
  long long f = (mm - (tr * K - tr + 1) + 1) % mod;
  long long n = tm - tr;
  long long ans3 = cal0(f, n, mod - (K - 1));
  return (ans1 + ans2 + ans3) % mod;
}
int main() {
  int i, n, K;
  cin >> n >> K;
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    while (top && a[i] > a[sta[top]]) r[sta[top]] = i - 1, top--;
    l[i] = sta[top] + 1, sta[++top] = i;
  }
  for (i = 1; i <= top; i++) r[sta[i]] = n;
  long long ans = 0;
  for (i = 1; i <= n; i++) add(ans, cal(l[i], r[i], i, K) * a[i] % mod);
  cout << ans << endl;
  return 0;
}

```