Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one. 

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000]. 

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead. 

Sample Input

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

很简单的题目，是一道KMP的模板题，但也是Hash的模板题。

Hash代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#include <unordered_map>
#define endl '\n'
#define sc(x) scanf("%d",&x)

using namespace std;
typedef unsigned long long ULL;
const int size=1e6+5;
ULL H[size],P[size];
int s[size];
void Hash(int len,int PA=2333)
{
	H[0]=0;
	P[0]=1;
	for(int i=1;i<=len;i++)
	{
		H[i]=H[i-1]*PA+s[i]+2000000;
		P[i]=P[i-1]*PA;
	}
}
ULL code(int l,int r)
{
	return H[r]-H[l-1]*P[r-l+1];
}
unordered_map<ULL,bool> mp;

int main()
{
	int t;
	sc(t);
	while(t--)
	{
		int n,m;
		sc(n),sc(m);
		for(int i=1;i<=n;i++) sc(s[i]);
		Hash(n);
		ULL s2=0;
		int t;
		for(int i=0;i<m;i++)
		{
			sc(t);
			t+=2000000;
			s2*=2333;
			s2+=t;
		}
		int ans=0;
		int i;
		for(i=1;i+m-1<=n;i++)
		{
			if(code(i,i+m-1)==s2) 
			{
				ans=1;
				break;
			}
		}
		if(ans) cout<<i<<endl;
		else cout<<"-1"<<endl;
	}
	return 0;
}

KMP代码：

#include<algorithm>
#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
#define endl '\n'
#define sc(x) scanf("%d",&x)

using namespace std;
int nxt[10005];
int ss[1000005],s[10005]; 
int n,m;
void getnext()
{
	nxt[0]=0,nxt[1]=0;
	int j=0;
	for(int i=1;i<m;i++)
	{
		j=nxt[i];
		while(j&&s[i]!=s[j]) j=nxt[j];
		if(s[i]==s[j]) nxt[i+1]=j+1;
		else nxt[i+1]=0;
	}
}
int query()
{
	getnext();
	int j=0;
	int ans=0;
	for(int i=0;i<n;i++)
	{
		while(j&&ss[i]!=s[j]) j=nxt[j];
		if(s[j]==ss[i]) ++j;
		if(j==m) return i-m+2;
	}
	return -1;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		memset(nxt,0,sizeof(nxt));
		sc(n),sc(m);
		for(int i=0;i<n;i++) sc(ss[i]);
		for(int i=0;i<m;i++) sc(s[i]);
		
		int ans=query();
		cout<<ans<<endl;
	}
	return 0;
}