#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
long long arr[1000000], L[1000000], R[1000000], st[1000000], k;
const long long MOD = 1e9 + 7;
inline long long f(long long L) {
  long long t = L / k;
  if (t == 0) return 0;
  if (L % k == 0) --t;
  if (t == 0) return 0;
  return t * L - k * (t + 1) * t / 2;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n, top, l, r, ans = 0, tmp;
  cin >> n >> k, --k;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  top = -1;
  for (int i = 0; i < n; ++i) {
    while (top > -1 && arr[i] >= arr[st[top]]) --top;
    if (top > -1)
      L[i] = st[top] + 1;
    else
      L[i] = 0;
    st[++top] = i;
  }
  top = -1;
  for (int i = n - 1; i >= 0; --i) {
    while (top > -1 && arr[i] > arr[st[top]]) --top;
    if (top > -1)
      R[i] = st[top] - 1;
    else
      R[i] = n - 1;
    st[++top] = i;
  }
  for (int i = 0; i < n; ++i) {
    tmp = ans;
    ans = (ans + f(R[i] - L[i] + 1) % MOD * arr[i]) % MOD;
    ans = (ans - f(i - L[i]) % MOD * arr[i] % MOD + MOD) % MOD;
    ans = (ans - f(R[i] - i) % MOD * arr[i] % MOD + MOD) % MOD;
  }
  cout << ans << "\n";
}

### Prompt

Please provide a cpp coded solution 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>
#pragma GCC optimize("Ofast")
using namespace std;
long long arr[1000000], L[1000000], R[1000000], st[1000000], k;
const long long MOD = 1e9 + 7;
inline long long f(long long L) {
  long long t = L / k;
  if (t == 0) return 0;
  if (L % k == 0) --t;
  if (t == 0) return 0;
  return t * L - k * (t + 1) * t / 2;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n, top, l, r, ans = 0, tmp;
  cin >> n >> k, --k;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  top = -1;
  for (int i = 0; i < n; ++i) {
    while (top > -1 && arr[i] >= arr[st[top]]) --top;
    if (top > -1)
      L[i] = st[top] + 1;
    else
      L[i] = 0;
    st[++top] = i;
  }
  top = -1;
  for (int i = n - 1; i >= 0; --i) {
    while (top > -1 && arr[i] > arr[st[top]]) --top;
    if (top > -1)
      R[i] = st[top] - 1;
    else
      R[i] = n - 1;
    st[++top] = i;
  }
  for (int i = 0; i < n; ++i) {
    tmp = ans;
    ans = (ans + f(R[i] - L[i] + 1) % MOD * arr[i]) % MOD;
    ans = (ans - f(i - L[i]) % MOD * arr[i] % MOD + MOD) % MOD;
    ans = (ans - f(R[i] - i) % MOD * arr[i] % MOD + MOD) % MOD;
  }
  cout << ans << "\n";
}

```