Number Sequence

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

Sample Output

6-1

#include<cstdio>#include<cstring>int a[1000005],b[1000005],next[1000005];void makenext(int *b,int m){next[0]=0;int k=0;for(int i=1;i<m;i++){while(k>0&&b[k]!=b[i])k=next[k-1];if(b[i]==b[k])k++;next[i]=k;}}int kmp(int *a,int n,int *b,int m){int k=0,i;for(i=0;i<n;i++){while(k>0&&a[i]!=b[k])k=next[k-1];if(b[k]==a[i])k++;if(k==m)return i-m+1+1;}if(i==n&&k<m)return -1;}int main(){int T,n,m,i,j,l1,l2,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]);makenext(b,m);t=kmp(a,n,b,m);printf("%d\n",t);}return 0;}