A - Number Sequence HDU - 1711

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
#include
#include
#include

using namespace std;

void ne(int *next, int *a, int n)
{
    next[0] = -1;
    int k = -1;
    int i = 0;
    while(i < n - 1)
    {
        if(k == -1 || a[i] == a[k])
        {
            next[++i] = ++k;
        }
        else {
            k = next[k];
        }
    }
}

int pre(int *a, int *b, int n, int m)
{
    int next[10010];
    ne(next, b, m);
    int i = 0;
    int j = 0;
    int flag = -1;
    while(i < n && j < m)
    {
        if(a[i] == b[j] || j == -1)
        {
            i++;
            j++;
        }
        else
        {
            j = next[j];
        }
        if(j == m){
            flag = i - j + 1;
            break;
        }
    }
    return flag;
}
int a[1001000], b[10010];
int main()
{
    int t;
    scanf("%d", &t);
    int n, m;
    while(t--){
        scanf("%d %d", &n, &m);
        for(int i = 0; i <= n - 1; i++){
            scanf("%d", &a[i]);
        }
        for(int i = 0; i <= m - 1; i++){
            scanf("%d", &b[i]);
        }
        printf("%d\n", pre(a, b, n, m));
    }
    return 0;
}