Problem Description

Victor loves to play with string. He thinks a string is charming as the string is a palindromic string.



Victor wants to play n times.
Each time he will do one of following four operations.



Operation 1 : add a char c to
the beginning of the string.



Operation 2 : add a char c to
the end of the string.



Operation 3 : ask the number of different charming substrings.



Operation 4 : ask the number of charming substrings, the same substrings which starts in different location has to be counted.



At the beginning, Victor has an empty string.

 

Input

The input contains several test cases, at most 5 cases.



In each case, the first line has one integer n means
the number of operations.



The first number of next n line
is the integer op,
meaning the type of operation. If op=1 or 2,
there will be a lowercase English letters followed.



1≤n≤100000.

 

Output

For each query operation(operation 3 or 4), print the correct answer.

 

Sample Input

6

1 a

1 b

2 a

2 c

3

4

8

1 a

2 a

2 a

1 a

3

1 b

3

4

 

Sample Output

4

5

4

5

11


这道题目和简单的回文树应用不一样了，它是在两头加字符，不是从左到右只加一边。那么
我们怎么解决这样的问题呢？
首先s[]数组应该开到长度的两倍，然后以中间作为起始点，分别向两边扩展。
然后last应该有两个指针，一个指向左边，一个指向右边
另外也应该有两个指针表示两边数组的长度，距离中间的长度
在加入新节点的时候，从左边插，和从右边插是一样的，只是有一点
在插入一边的时候可能会应该回影响另一边的last指针。具体见代码

#include <iostream>

#include <string.h>

#include <stdlib.h>

#include <algorithm>

#include <math.h>

#include <stdio.h>

using namespace std;

typedef long long int LL;

const int MAX=1e5+5;

char str[MAX];

int n;

struct Tree

{

    int next[2*MAX][26];

    int fail[2*MAX];

    int num[2*MAX];

    int len[2*MAX];

    int s[2*MAX];

    int last[2];

    int tot[2];

    LL p;

    int new_node(int x)

    {

        memset(next[p],0,sizeof(next[p]));

        num[p]=0;

        len[p]=x;

        return p++;

    }

    void init()

    {

        p=0;

        new_node(0);

        new_node(-1);

        last[0]=0;

        last[1]=0;

        tot[0]=MAX-5;

        tot[1]=MAX-6;

        fail[0]=1;

    }

    int get_fail(int x,int k)

    {

        s[tot[0]-1]=-1;s[tot[1]+1]=-1;

         while(s[tot[k]]!=s[tot[k]+(k?-1:1)*(len[x]+1)])

            x=fail[x];

         return x;

    }

    int add(int x,int k)

    {

        x-='a';

        s[tot[k]+=(k?1:-1)]=x;

        int cur=get_fail(last[k],k);

        if(!(last[k]=next[cur][x]))

        {

            int now=new_node(len[cur]+2);

            fail[now]=next[get_fail(fail[cur],k)][x];

            next[cur][x]=now;

            num[now]=num[fail[now]]+1;

            last[k]=now;

            if(len[last[k]]==tot[1]-tot[0]+1)  last[k^1]=last[k];

        }

        return num[last[k]];

    }

}tree;

int main()

{

    int x;char y[10];

    while(scanf("%d",&n)!=EOF)

    {

        tree.init();

        LL ans=0;

        for(int i=1;i<=n;i++)

        {

            scanf("%d",&x);

            if(x==1)

            {

                scanf("%s",y);

                ans+=tree.add(y[0],0);

            }

            else if(x==2)

            {

                scanf("%s",y);

                ans+=tree.add(y[0],1);

            }

            else if(x==3)

                printf("%d\n",tree.p-2);

            else

                printf("%lld\n",ans);

        }

    }

    return 0;

}