CodeForces 367B Sereja ans Anagrams

Description

Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, …, an. Similarly, sequence b consists of m integers b1, b2, …, bm. 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 aq, aq + p, aq + 2p, …, aq + (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, …, an(1 ≤ ai ≤ 109). The next line contains m integers b1, b2, …, bm(1 ≤ bi ≤ 109).

Output

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

Sample Input

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

思路：

首先map可以直接用等号来判断两个map是否一样。

其次，我们枚举起点，而且外层只用从1枚举到p,内后像队列一样删掉对头，队尾插入来维护map，即可；

#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;
int n,m,p;
int cnt=0,q;
int a[2000090];
map<int,int> tmp1,tmp2;
int ans[2000090];
int main()
{
scanf("%d%d%d",&n,&m,&p);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
int t;
for(int i=0;i<m;i++)
{
scanf("%d",&t);
tmp1[t]++;
}
for(int i=1;i<=p;i++)
{
tmp2.clear();
for(long long j=0;j<m&&i+j*p<=n;j++)
tmp2[a[i+j*p]]++;
if(tmp1==tmp2)
ans[cnt++]=i;
for(long long j=m;i+j*p<=n;j++)
{
int pre=a[i+(j-m)*p];
if(tmp2[pre]==1)
tmp2.erase(pre);
else tmp2[pre]--;
tmp2[a[i+j*p]]++;
if(tmp2==tmp1)
ans[cnt++]=i+(j-m+1)*p;
}
}
printf("%d\n",cnt);
sort(ans,ans+cnt);
for(int i=0;i<cnt;i++)
{
if(i) printf(" ");
printf("%d",ans[i]);
}
puts("");
return 0;
}