#include <bits/stdc++.h>
using namespace std;
const int N = 1000000 + 10;
const long long MOD = 1e9 + 7;
int n, k;
int a[N], pre[N], nex[N];
int stk[N], top;
long long getAns(long long x) {
  if (x < k) return 0;
  long long c = (x - k) / (k - 1) + 1;
  long long a1 = (x - k + 1);
  long long len = (x - k) / (k - 1) * (k - 1) + k;
  long long ac = x - len + 1;
  return c * (a1 + ac) / 2;
}
long long solve(int pos, int L, int R) {
  return getAns(R - L - 1) - getAns(pos - L - 1) - getAns(R - pos - 1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    while (top && a[stk[top]] <= a[i]) top--;
    pre[i] = stk[top], stk[++top] = i;
  }
  top = 0;
  stk[top] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (top && a[stk[top]] < a[i]) top--;
    nex[i] = stk[top], stk[++top] = i;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = ans + solve(i, pre[i], nex[i]) % MOD * a[i] % MOD;
    ans = ans % MOD;
  }
  cout << ans << endl;
}

### 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 = 1000000 + 10;
const long long MOD = 1e9 + 7;
int n, k;
int a[N], pre[N], nex[N];
int stk[N], top;
long long getAns(long long x) {
  if (x < k) return 0;
  long long c = (x - k) / (k - 1) + 1;
  long long a1 = (x - k + 1);
  long long len = (x - k) / (k - 1) * (k - 1) + k;
  long long ac = x - len + 1;
  return c * (a1 + ac) / 2;
}
long long solve(int pos, int L, int R) {
  return getAns(R - L - 1) - getAns(pos - L - 1) - getAns(R - pos - 1);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    while (top && a[stk[top]] <= a[i]) top--;
    pre[i] = stk[top], stk[++top] = i;
  }
  top = 0;
  stk[top] = n + 1;
  for (int i = n; i >= 1; i--) {
    while (top && a[stk[top]] < a[i]) top--;
    nex[i] = stk[top], stk[++top] = i;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    ans = ans + solve(i, pre[i], nex[i]) % MOD * a[i] % MOD;
    ans = ans % MOD;
  }
  cout << ans << endl;
}

```