Problem Description

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>

int n, m;
int a[1000005];
int b[10005];
int f[10005];

void getfail()
{
    f[0] = 0;
    f[1] = 0;
    for (int i = 1; i < m; i++)
    {
        int j = f[i];
        while (j && b[i] != b[j])
            j = f[j];
        f[i+1] = b[j] == b[i] ? j + 1 : 0;
    }
}

int main()
{
    int flag, i, j, t;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d %d",&n,&m);
        for (i = 0; i < n; i++)
            scanf("%d",&a[i]);
        for (i = 0; i < m; i++)
            scanf("%d",&b[i]);
        getfail();
        for (i = 0;i <= m; i++) printf("%d ",f[i]);  puts("");
        flag = 1;
        j = 0;
        for (i = 0;i < n;i++)
        {
            while (j && b[j] != a[i])
                j = f[j];
            if (b[j] == a[i])
                j++;
            if (j == m)
            {
                flag = 0;
                printf("%d\n",i - m + 1);
            }
        }
        if (flag)
            puts("-1");
    }
    return 0;
}