#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000111;
const int MOD = 1000000007;
int sr;
int sum(int a, int b) {
  sr = a + b;
  if (sr >= MOD) sr -= MOD;
  return sr;
}
int mul(int a, int b) { return (int)((1LL * a * b) % (1LL * MOD)); }
int N, K;
int A[MAXN], L[MAXN], R[MAXN];
int St[MAXN], Top;
int C[MAXN];
int Calc(int p, int l, int r) {
  l -= (p - 1);
  r -= (p - 1);
  return C[r] - C[l - 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> K;
  --K;
  for (int i = K + 1; i <= N; i += K) C[i] = 1;
  for (int i = 1; i <= N; ++i) C[i] += C[i - 1];
  for (int i = 1; i <= N; ++i) cin >> A[i];
  Top = 0;
  St[0] = 0;
  for (int i = 1; i <= N; ++i) {
    while (Top > 0 && A[St[Top]] < A[i]) --Top;
    L[i] = St[Top] + 1;
    St[++Top] = i;
  }
  Top = 0;
  St[0] = N + 1;
  for (int i = N; i >= 1; --i) {
    while (Top > 0 && A[St[Top]] <= A[i]) --Top;
    R[i] = St[Top] - 1;
    St[++Top] = i;
  }
  int Ans = 0;
  for (int i = 1; i <= N; ++i) {
    if (R[i] - i > i - L[i]) {
      for (int j = L[i]; j <= i; ++j) {
        Ans = sum(Ans, mul(Calc(j, i, R[i]), A[i]));
      }
    } else {
      for (int j = i; j <= R[i]; ++j) {
        Ans = sum(Ans, mul(Calc(N + 1 - j, N + 1 - i, N + 1 - L[i]), A[i]));
      }
    }
  }
  cout << Ans << endl;
  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;
const int MAXN = 1000111;
const int MOD = 1000000007;
int sr;
int sum(int a, int b) {
  sr = a + b;
  if (sr >= MOD) sr -= MOD;
  return sr;
}
int mul(int a, int b) { return (int)((1LL * a * b) % (1LL * MOD)); }
int N, K;
int A[MAXN], L[MAXN], R[MAXN];
int St[MAXN], Top;
int C[MAXN];
int Calc(int p, int l, int r) {
  l -= (p - 1);
  r -= (p - 1);
  return C[r] - C[l - 1];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> N >> K;
  --K;
  for (int i = K + 1; i <= N; i += K) C[i] = 1;
  for (int i = 1; i <= N; ++i) C[i] += C[i - 1];
  for (int i = 1; i <= N; ++i) cin >> A[i];
  Top = 0;
  St[0] = 0;
  for (int i = 1; i <= N; ++i) {
    while (Top > 0 && A[St[Top]] < A[i]) --Top;
    L[i] = St[Top] + 1;
    St[++Top] = i;
  }
  Top = 0;
  St[0] = N + 1;
  for (int i = N; i >= 1; --i) {
    while (Top > 0 && A[St[Top]] <= A[i]) --Top;
    R[i] = St[Top] - 1;
    St[++Top] = i;
  }
  int Ans = 0;
  for (int i = 1; i <= N; ++i) {
    if (R[i] - i > i - L[i]) {
      for (int j = L[i]; j <= i; ++j) {
        Ans = sum(Ans, mul(Calc(j, i, R[i]), A[i]));
      }
    } else {
      for (int j = i; j <= R[i]; ++j) {
        Ans = sum(Ans, mul(Calc(N + 1 - j, N + 1 - i, N + 1 - L[i]), A[i]));
      }
    }
  }
  cout << Ans << endl;
  return 0;
}

```