Sereja ans Anagrams 优先队列控制

Sereja has two sequences a and b and number p. Sequence a consists of n integers a₁, a₂, ..., a_n. Similarly, sequence b consists of m integers b₁, b₂, ..., b_m. As usual, Sereja studies the sequences he has. Today he wants to find the number of positions q (q + (m - 1)·p ≤ n; q ≥ 1), such that sequence b can be obtained from sequence a_q, a_q + p, a_q + 2p, ..., a_{q + (m - 1)p} by rearranging elements.

Sereja needs to rush to the gym, so he asked to find all the described positions of q.

Input

The first line contains three integers n, m and p (1 ≤ n, m ≤ 2·10⁵, 1 ≤ p ≤ 2·10⁵). The next line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹). The next line contains m integers b₁, b₂, ..., b_m (1 ≤ b_i ≤ 10⁹).

Output

In the first line print the number of valid qs. In the second line, print the valid values in the increasing order.

Example

Input

5 3 1
1 2 3 2 1
1 2 3

Output

2
1 3

Input

6 3 2
1 3 2 2 3 1
1 2 3

Output

2
1 2


暴力肯定超时，总算写出来了，想到了只需要满足当判断了m个元素满足后，删除最初的那个元素以及标记，然后接着往下判断，但是写的很麻烦，后来用优先队列，但是又出了小问题（代码中有表示），很小的问题，以至于我烦的找不到。


代码：

#include <iostream>
#include <cstdio>
#include <queue>
#include <map>
#include <algorithm>
using namespace std;
struct que
{
    int id,num;
    friend bool operator<(que a,que b)
    {
        return a.id>b.id;
    }
}t;
int n,m,p,a[200000],b[200000],k=0,c,ans[200000];
int main()
{
    map<int,int> mark;
    scanf("%d%d%d",&n,&m,&p);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(int i=0;i<m;i++)
    {
        scanf("%d",&b[i]);
        mark[b[i]]++;
    }
    for(int i=0;i<p;i++)
    {
        map<int,int>check;
        priority_queue <que> q;
        for(int j=i;j<n;j+=p)
        {
            //cout<<q.size()<<endl;
            if(mark[a[j]]==0)
            {
                while(!q.empty()){check[a[q.top().num]]--;q.pop();}///check的下标应该是a数组的元素 而不是a的下标
                continue;
            }
            if(check[a[j]]<mark[a[j]])
            {
                check[a[j]]++;
                if(q.empty())
                {
                    t.id=1,t.num=j;
                    q.push(t);
                }
                else
                {
                    t.id=q.top().id+q.size();
                    t.num=j;
                    q.push(t);
                }
            }
            else
            {
                while(!q.empty()&&a[q.top().num]!=a[j])///判断 对应位置的元素是否相等，而我当时判断了下标，傻逼吧，下标怎么可能相等，热傻了吧
                {
                    check[a[q.top().num]]--;
                    q.pop();
                }
                t.id=q.top().id+q.size();
                t.num=j;
                q.push(t);
                q.pop();
            }
            if(q.size()==m)
            {
                //cout<<q.top().num<<endl;
                ans[k++]=q.top().num+1;
                check[a[q.top().num]]--;
                q.pop();
                //cout<<q.top().num<<endl;
            }
        }
    }
    printf("%d\n",k);
    sort(ans,ans+k);
    for(int i=0;i<k;i++)
        printf("%d ",ans[i]);
}