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 mintegers 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.

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).

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

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

using namespace std;

int n,m,p,total,flag,now;

int a[],b[],c[],all[],visited[];

void hebing()//合并成一个数组，全部存在c数组里

{

    int i,j,x;

    total=;

    for(i=;i<=n;i++)

     c[++total]=a[i];

    for(i=;i<=m;i++)

     c[++total]=b[i];

    sort(c+,c++total);

    x=;

    for(i=;i<=total;i++)//去重

     if(i==||c[i]!=c[i-])

      c[++x]=c[i];

    total=x;

}

int find(int x)//二分查找x处在c数组中的位置

{

    int l=,r=total+,mid;

    while(l<=r)

    {

        mid=(l+r)>>;

        if(c[mid]==x)

         return mid;

        else if(c[mid]>x)

         r=mid-;

        else

         l=mid+;

    }

}

void lisanhua()//对a数组和b数组进行离散化

{

    int i;

    for(i=;i<=n;i++)

     a[i]=find(a[i]);

    for(i=;i<=m;i++)

     b[i]=find(b[i]);

}

void init()

{

    int i;

    flag=;

    memset(all,,sizeof(all));

    for(i=;i<=m;i++)//哈希处理b数组的情况

    {

        if(all[b[i]]==)

         flag++;

        all[b[i]]++;

    }

}

void add(int x)

{

    visited[x]++;

    if(visited[x]==all[x])

     now++;

}

void del(int x)

{

    visited[x]--;

    if(visited[x]+==all[x])

     now--;

}

void solve()

{

    int res[],num=,i,j,k;

    for(i=;i<=p;i++)//因为相隔为p，所以只需枚举1-p，然后对每一种序列滚动过去

    {

        if((i+(long long)(m-)*p)<=n)//因为(m-1)*p已经超过int型，所以要强制转换成long long

        {                            //否则就会出现Runtime Error

            now=;

            for(j=;j<m;j++)

             add(a[i+j*p]);

            if(now==flag)

             res[num++]=i;

            for(j=i+p;(j+(long long)(m-)*p)<=n;j=j+p)//滚动过去

            {

                del(a[j-p]);

                add(a[j+(m-)*p]);

                if(now==flag)

                 res[num++]=j;

            }

            for(k=;k<m;k++)//把加了的删掉，开始我用了memset(visited,0,sizeof(visited));

             del(a[j-p+k*p]);//然后果断超时了

        }

    }

    printf("%d\n",num);

    sort(res,res+num);

    for(i=;i<num;i++)

     if(i!=num-)

      printf("%d ",res[i]);

     else

      printf("%d\n",res[i]);

}

int main()

{

    int i,j;

    scanf("%d%d%d",&n,&m,&p);

    {

        for(i=;i<=n;i++)

         scanf("%d",&a[i]);

        for(i=;i<=m;i++)

         scanf("%d",&b[i]);

        hebing();

        lisanhua();

        init();

        solve();

    }

    return ;

}