Problem Description

Hotaru Ichijou recently is addicated to math problems. Now she is playing with N-sequence.

Let's define N-sequence, which is composed with three parts and satisfied with the following condition:

1. the first part is the same as the thrid part,

2. the first part and the second part are symmetrical.

for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.



Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.

 

Input

There are multiple test cases. The first line of input contains an integer T（T<=20）, indicating the number of test cases. 



For each test case:



the first line of input contains a positive integer N（1<=N<=100000）, the length of a given sequence



the second line includes N non-negative integers ,each interger is no larger than 109 ,
descripting a sequence.

 

Output

Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by a integer, the largest length of N-sequence.



We guarantee that the sum of all answers is less than 800000.

 

Sample Input

1

10

2 3 4 4 3 2 2 3 4 4

 

Sample Output

Case #1: 9

这题能够用Manacher算法做。由于题目要找的是三段（第一段和第二段对称，第二段和第三段对称）。事实上就是两个连在一起的回文串，我们能够先用Manacher算法初始化各个点的p[i]值（即能够向右延伸的最大距离。包含本身，这时已经增加了-1取代算法中的'#',-2取代算法中的'$'），然后对于每一个i。枚举j（j属于1~p[i]-1),假设i+j-p[i+j]+1<=i，那么说明i。j能够分别作为第一、二段的点和第二、三段的点）。

这里有个优化，由于枚举时满足条件的仅仅有'#'（即'-1’），所以我们能够使i，j每次变化2.
#include<iostream>

#include<stdio.h>

#include<stdlib.h>

#include<string.h>

#include<math.h>

#include<vector>

#include<map>

#include<set>

#include<queue>

#include<stack>

#include<string>

#include<algorithm>

using namespace std;

#define maxn 100060

int a[maxn],b[2*maxn],p[2*maxn];

int main()

{

    int n,m,i,j,T,mx,idx,maxx,num1=0;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d",&n);

        for(i=0;i<n;i++){

            scanf("%d",&a[i]);

        }

        if(n<3){

            printf("0\n");continue;

        }

        b[0]=-2;

        b[1]=-1;

        for(i=0;i<n;i++){

            b[i*2+2]=a[i];

            b[i*2+3]=-1;

        }

        n=2*n+2;mx=0;

        for(i=0;i<n;i++){

            if(i<mx){

                p[i]=min(p[idx*2-i],mx-i);

            }

            else p[i]=1;

            while(b[i-p[i]]==b[i+p[i]]){

                p[i]++;

            }

            if(mx<i+p[i]){

                mx=i+p[i];

                idx=i;

            }

        }

        maxx=0;

        for(i=3;i<n;i+=2){

            for(j=p[i]-1;j>=1;j-=2){

                if(j<maxx)break;

                if(i+j-p[i+j]+1<=i){

                    maxx=max(maxx,j);break;

                }

            }

        }

        num1++;

        printf("Case #%d: ",num1);

        printf("%d\n",maxx/2*3);

    }

    return 0;

}