#include <bits/stdc++.h>
using namespace std;
const long double Pi = 3.14159265359;
const long long MOD = (long long)1e9 + 7;
const long long MAXN = (long long)1e6 + 10;
const long long INF = (long long)9223372036854775;
const long double EPS = (long double)1e-8;
long long dp[MAXN];
long long dpx[MAXN];
long long a[MAXN];
deque<long long> de;
stack<long long> s;
long long mk[MAXN];
long long lb[MAXN];
int main() {
  long long n, k;
  scanf("%lld", &n);
  scanf("%lld", &k);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < k - 1; i++) {
    while (de.size() != 0) {
      if (a[de.back()] <= a[i])
        de.pop_back();
      else
        break;
    }
    de.push_back(i);
  }
  for (int i = k - 1; i < n; i++) {
    while (de.size() != 0) {
      if (a[de.back()] <= a[i])
        de.pop_back();
      else
        break;
    }
    de.push_back(i);
    if (i - de.front() >= k) de.pop_front();
    mk[i] = de.front();
  }
  de.clear();
  for (int i = 0; i < n; i++) {
    while (de.size() != 0) {
      if (a[de.back()] <= a[i])
        de.pop_back();
      else
        break;
    }
    if (de.size() == 0)
      lb[i] = -1;
    else
      lb[i] = de.back();
    de.push_back(i);
  }
  long long nex, cnt, pa;
  for (int i = k - 1; i < n; i++) {
    nex = lb[mk[i]];
    if (nex == -1) {
      dpx[i] = ((a[mk[i]]) * (i / (k - 1))) % MOD;
    } else {
      cnt = i - nex - 1;
      pa = cnt / (k - 1);
      dpx[i] = (dpx[i - (pa * (k - 1))] + ((a[mk[i]]) * pa)) % MOD;
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) ans = (ans + dpx[i]) % MOD;
  printf("%lld", ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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>
using namespace std;
const long double Pi = 3.14159265359;
const long long MOD = (long long)1e9 + 7;
const long long MAXN = (long long)1e6 + 10;
const long long INF = (long long)9223372036854775;
const long double EPS = (long double)1e-8;
long long dp[MAXN];
long long dpx[MAXN];
long long a[MAXN];
deque<long long> de;
stack<long long> s;
long long mk[MAXN];
long long lb[MAXN];
int main() {
  long long n, k;
  scanf("%lld", &n);
  scanf("%lld", &k);
  for (int i = 0; i < n; i++) scanf("%lld", &a[i]);
  for (int i = 0; i < k - 1; i++) {
    while (de.size() != 0) {
      if (a[de.back()] <= a[i])
        de.pop_back();
      else
        break;
    }
    de.push_back(i);
  }
  for (int i = k - 1; i < n; i++) {
    while (de.size() != 0) {
      if (a[de.back()] <= a[i])
        de.pop_back();
      else
        break;
    }
    de.push_back(i);
    if (i - de.front() >= k) de.pop_front();
    mk[i] = de.front();
  }
  de.clear();
  for (int i = 0; i < n; i++) {
    while (de.size() != 0) {
      if (a[de.back()] <= a[i])
        de.pop_back();
      else
        break;
    }
    if (de.size() == 0)
      lb[i] = -1;
    else
      lb[i] = de.back();
    de.push_back(i);
  }
  long long nex, cnt, pa;
  for (int i = k - 1; i < n; i++) {
    nex = lb[mk[i]];
    if (nex == -1) {
      dpx[i] = ((a[mk[i]]) * (i / (k - 1))) % MOD;
    } else {
      cnt = i - nex - 1;
      pa = cnt / (k - 1);
      dpx[i] = (dpx[i - (pa * (k - 1))] + ((a[mk[i]]) * pa)) % MOD;
    }
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) ans = (ans + dpx[i]) % MOD;
  printf("%lld", ans);
  return 0;
}

```