Number Sequence HDU

Given two sequences of numbers : a11 , a22 , ...... , aNN , and b11 , b22 , ...... , bMM (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make aKK = b11 , aK+1K+1 = b22 , ...... , aK+M−1K+M−1 = bMM . If there are more than one K exist, output the smallest one.
InputThe 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 a11 , a22 , ...... , aNN . The third line contains M integers which indicate b11 , b22 , ...... , bMM . All integers are in the range of −1000000,1000000−1000000,1000000 .
OutputFor 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<iostream>
#include<cstring>
#include<string>
#include<cstdio>
using namespace std;

int nxt[1000005],s[1000005],p[10005]; 
int T,n,m;

void getNext(){
    nxt[0]=-1;
    for(int i=0;i<m;i++){
        int k=nxt[i];
        while(k!=-1&&p[i]!=p[k]) k=nxt[k];
        nxt[i+1]=k+1;
    }
}

int kmp(){
    getNext();
    int i=0,j=0;
    while(i<n&&j<m){
        while(j!=-1&&s[i]!=p[j]) j=nxt[j];
        ++i;
        ++j;   
    }
    if(j==m) return i-j+1;
    return -1;
}

int main()
{   scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++) scanf("%d",&s[i]);
        for(int i=0;i<m;i++) scanf("%d",&p[i]);
        int ans=kmp();
        printf("%d
",ans);
    }
    return 0;
}