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

The main idea of ​​the topic: Find a continuous subsequence in sequence a that is equal to sequence b, and output the earliest one, if not, output -1.

Problem-solving ideas: Just set the KMP board. PS: I wrote a lot last night, but suddenly the network card returned, and when I came back in, it was gone, so I didn't write.

You can read this blog:KMP algorithm (research summary, string)

Code:

#pragma GCC optimize(2)
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <stack>
#include <bitset>
#include <queue>
//#include <random>
#include <time.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ls root<<1
#define rs root<<1|1
const int maxn=1e4+10;
const int maxm=1e6+10;
//std::mt19937 rnd(time(NULL));
int to[maxn],a[maxm],b[maxn],n,m;
void init()
{
    to[1]=0;
    for(int i=2;i<=m;i++){
        int t=to[i-1];
        while(t && b[t+1]!=b[i])t=to[t];
        if(b[t+1]==b[i])to[i]=t+1;
        else to[i]=0;
    }
}

int kmp()
{
    init();
    int i=1,j=1;
    while(i<=n){
        if(a[i]==b[j]){
            i++,j++;
            if(j==m+1){
                return i-m;
            }
        }
        else{
            if(j==1) i++;
            else j=to[j-1]+1;
        }
    }
    return -1;
}
signed main()
{
    int t;
    scanf("%lld",&t);
    while(t--){
        scanf("%lld%lld",&n,&m);
        for(int i=1;i<=n;i++){
            scanf("%lld",a+i);
        }
        for(int i=1;i<=m;i++)scanf("%lld",b+i);
        printf("%lld\n",kmp());
    }
}