#include <bits/stdc++.h>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:3336777216")
using namespace std;
int IT_MAX = 1 << 19;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
set<int> Sx;
long long in[1000050];
long long u[1000050];
long long getcnt(int L, int K) {
  if (L < K) return 0;
  long long mx = (L - 1) / (K - 1);
  return L * mx - (K - 1) * mx * (mx + 1) / 2;
}
int main() {
  int N, K, i, j;
  scanf("%d %d", &N, &K);
  for (i = 1; i <= N; i++) {
    scanf("%lld", &in[i]);
    u[i] = i;
  }
  sort(u + 1, u + N + 1,
       [](long long a, long long b) { return in[a] > in[b]; });
  Sx.insert(0);
  Sx.insert(N + 1);
  long long ans = 0;
  for (i = 1; i <= N; i++) {
    int p = u[i];
    auto it = Sx.lower_bound(p);
    int S = *prev(it), E = *it;
    long long c =
        getcnt(E - S - 1, K) - getcnt(p - S - 1, K) - getcnt(E - p - 1, K);
    c %= MOD;
    ans = (ans + c * in[p]) % MOD;
    Sx.insert(p);
  }
  return !printf("%lld\n", ans);
}

### 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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:3336777216")
using namespace std;
int IT_MAX = 1 << 19;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
set<int> Sx;
long long in[1000050];
long long u[1000050];
long long getcnt(int L, int K) {
  if (L < K) return 0;
  long long mx = (L - 1) / (K - 1);
  return L * mx - (K - 1) * mx * (mx + 1) / 2;
}
int main() {
  int N, K, i, j;
  scanf("%d %d", &N, &K);
  for (i = 1; i <= N; i++) {
    scanf("%lld", &in[i]);
    u[i] = i;
  }
  sort(u + 1, u + N + 1,
       [](long long a, long long b) { return in[a] > in[b]; });
  Sx.insert(0);
  Sx.insert(N + 1);
  long long ans = 0;
  for (i = 1; i <= N; i++) {
    int p = u[i];
    auto it = Sx.lower_bound(p);
    int S = *prev(it), E = *it;
    long long c =
        getcnt(E - S - 1, K) - getcnt(p - S - 1, K) - getcnt(E - p - 1, K);
    c %= MOD;
    ans = (ans + c * in[p]) % MOD;
    Sx.insert(p);
  }
  return !printf("%lld\n", ans);
}

```