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

题意：

      给你两个数组a,b，问是否存在一个m使得a[m]=b[1]，a[m+1]=b[2].....。输出最小的m，如果不存在的话，输出-1。

思路：

     这题的意思其实就是判断在b串在a串的哪个位置出现，包含在a串中。就用kmp模板然后，当b串搜索结束后，输出a串中当前位置i - q(b串长度)+1即可（因为i是a串现在所处的位置，q为b串长度，我从0开始读入的，如果从i = 1读入就不需要加1了）。

代码：

#include<stdio.h>
#include<string.h>
int a[1000010],b[10010];
int next[10010];
int n,p,q;
void get_next()
{
	int i,j;
	i=1;
	j=0;
	while(i<q)
	{
		if(j==0&&b[i]!=b[j])
		{
			next[i]=0;
			i++;	
		}
		else if(j>0&&b[i]!=b[j])
		{
			j=next[j-1];
		}
		else
		{
			next[i]=j+1;
			i++;
			j++;
		}
	}
}
int kmp()
{
	int i,j,flag=0;
	i=0;
	j=0;
	while(i<p&&j<q)
	{
		if(j==0&&a[i]!=b[j])
		{
			i++;
		}
		else if(j>0&&a[i]!=b[j])
		{
			j=next[j-1];
		}
		else
		{
			i++;
			j++;
		}
	}
	if(j>=q)                       //当b串搜索完
		printf("%d\n",i+1-q);  //因为i是a串现在所处的位置，q为b串长度
	else
		printf("-1\n");
 } 
int main()
{
	int i;
	scanf("%d",&n);
	while(n--)
	{
		scanf("%d%d",&p,&q);
		for(i=0;i<p;i++)
			scanf("%d",&a[i]);
		for(i=0;i<q;i++)
			scanf("%d",&b[i]);
		get_next();
		kmp();
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
	}
}