Number Sequence --（kmp）

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

#include <stdio.h>
const int maxn=1e6+10;
int t,n,m,a[maxn],b[maxn],nxt[maxn];
void ans(){
	int i=1;
	int j=0;
	nxt[0]=0;
	while(i<m)
	{
		if(b[i]==b[j]) nxt[i++]=++j;
		else if(!j)  i++;
		else   j=nxt[j-1];
	}
}
int kmp()
{
	int i=0,j=0;
	while(i<n&&j<m)
	{
		if(a[i]==b[j])  i++,j++;
		else if(!j)  i++;
		else j=nxt[j-1];
	}
	if(j==m) return i-m+1;
	else return -1;
}
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		for(int i=0;i<n;i++)
		    scanf("%d",&a[i]);
		for(int i=0;i<m;i++)
		    scanf("%d",&b[i]);
		ans();
		printf("%d\n",kmp());
	}
	return 0;
}