class Solution_Case	//我的小习惯
{
	/*
	1.统计排名可以利用树状数组或者线段树，不用非要离散化（无法动态维护）
	查询对应下标的前缀和，就可以查出前面有几个数字，而且可以动态维护
	2.动态维护后子串的排名，需要删除前面的数，即add-1
	*/
};

#include<bits/stdc++.h>
#define lowbit(x) (x&(-x))
#define N 500005
using namespace std;

int n, m, s, cnt;
int A[N], B[N];
int Rank1[N], Rank2[N];
int Next[N], ans[N];

class BIT //树状数组，单点修改，前缀查询
{
  public:
	int bitSum[N];
	void Add(int x, int val)
	{
		for (; x <= s; x += lowbit(x)) bitSum[x] += val;
	}
	int Query(int x, int ans = 0)
	{
		for (; x; x -= lowbit(x)) ans += bitSum[x];
		return ans;
	}
	bool Check(int x, int y)
	{
		//这里是为了应对重复的值，查询前一位，确保地位完全相同
		//变相的相当于记录了与自己相同的有多少个（Rank1-Rank2）
		return (Query(y) == Rank1[x]) && (Query(y - 1) == Rank2[x]);
	}
} Bit;

void Make_Next()	//处理出next数组
{
	memset(Bit.bitSum, 0, sizeof(Bit.bitSum));
	for (int i = 2, j = 0; i <= m; ++i)
	{
		//printf("j=%d i=%d\n", j, i);
		Bit.Add(B[i], 1);
		while (j && !Bit.Check(j + 1, B[i])) //失配
		{
			//删掉前面的点，因为要维护后子串的排名
			//printf("k=%d~%d add-1\n", i - j, i - Next[j]);
			for (int k = i - j; k < i - Next[j]; ++k)  Bit.Add(B[k], -1);
			j = Next[j];
		}
		if (Bit.Check(j + 1, B[i]))  ++j;
		Next[i] = j;
	}
}

void Cul_Ans()	//计算答案
{
	memset(Bit.bitSum, 0, sizeof(Bit.bitSum));
	for (int i = 1, j = 0; i <= n; ++i)
	{
		Bit.Add(A[i], 1);
		while (j && !Bit.Check(j + 1, A[i])) //失配
		{
			for (int k = i - j; k < i - Next[j]; ++k)  Bit.Add(A[k], -1);	//删点
			j = Next[j];
		}
		if (Bit.Check(j + 1, A[i]))  ++j;

		if (j == m)	//整个串匹配完毕
		{
			ans[++cnt] = i - m + 1;

			//匹配上了以后要向前跳继续匹配，所以仍要删点
			//printf("k=%d~%d add-1\n", i - j, i - Next[j]);
			for (int k = i - j + 1; k <= i - Next[j]; ++k)  Bit.Add(A[k], -1);
			j = Next[j];
		}
	}
}

int main()
{
	scanf("%d%d%d", &n, &m, &s);
	for (int i = 1; i <= n; ++i)  scanf("%d", &A[i]);

	//维护出B的每个前缀中末尾的排名，在kmp时候要用
	for (int i = 1; i <= m; ++i)
	{
		scanf("%d", &B[i]), Bit.Add(B[i], 1);
		Rank1[i] = Bit.Query(B[i]), Rank2[i] = Bit.Query(B[i] - 1);
	}

	Make_Next();
	Cul_Ans();
	printf("%d\n", cnt);
	for (int i = 1; i <= cnt; ++i)  printf("%d\n", ans[i]);
	return 0;
}