#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
int n, k, cnt[maxn], a[maxn], L[maxn], R[maxn];
stack<pair<int, int> > stk;
int main() {
  scanf("%d", &n);
  ;
  scanf("%d", &k);
  ;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  ;
  stk.push({INF, 0});
  for (int i = 1; i <= n; i++) {
    while (stk.top().first < a[i]) stk.pop();
    L[i] = stk.top().second + 1;
    stk.push({a[i], i});
  }
  while (stk.size()) stk.pop();
  stk.push({INF, n + 1});
  for (int i = n; i >= 1; i--) {
    while (stk.top().first <= a[i]) stk.pop();
    R[i] = stk.top().second - 1;
    stk.push({a[i], i});
  }
  for (int i = 1; i <= n; i++) {
    cnt[i] = (cnt[i - 1] + (i - 1) / (k - 1)) % mod;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int t = (cnt[R[i] - L[i] + 1] - cnt[i - L[i]] - cnt[R[i] - i]) % mod;
    ans += 1LL * t * a[i] % mod;
  }
  printf("%I64d\n", (ans % mod + mod) % mod);
  return 0;
}

### Prompt

Please formulate a cpp solution to 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 INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;
int n, k, cnt[maxn], a[maxn], L[maxn], R[maxn];
stack<pair<int, int> > stk;
int main() {
  scanf("%d", &n);
  ;
  scanf("%d", &k);
  ;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  ;
  stk.push({INF, 0});
  for (int i = 1; i <= n; i++) {
    while (stk.top().first < a[i]) stk.pop();
    L[i] = stk.top().second + 1;
    stk.push({a[i], i});
  }
  while (stk.size()) stk.pop();
  stk.push({INF, n + 1});
  for (int i = n; i >= 1; i--) {
    while (stk.top().first <= a[i]) stk.pop();
    R[i] = stk.top().second - 1;
    stk.push({a[i], i});
  }
  for (int i = 1; i <= n; i++) {
    cnt[i] = (cnt[i - 1] + (i - 1) / (k - 1)) % mod;
  }
  long long ans = 0;
  for (int i = 1; i <= n; i++) {
    int t = (cnt[R[i] - L[i] + 1] - cnt[i - L[i]] - cnt[R[i] - i]) % mod;
    ans += 1LL * t * a[i] % mod;
  }
  printf("%I64d\n", (ans % mod + mod) % mod);
  return 0;
}

```