Barty have a computer, it can do these two things.

1. Add a new string to its memory, the length of this string is even.

2. For given $4$ strings $a,b,c,d$, find out how many strings that can be product by $a+s1+b+c+s2+d$, and $|a|+|s1|+|b|=|c|+|s2|+|d$

   ∣. $|s|$ means the length of string $s$, $s1$ and $s2$ can be any string, including "".

Please help your computer to do these things.

Input Format

Test cases begins with $T(T\le 5)$.

Then $T$ test cases follows.

Each test case begins with an integer $Q(Q\le 30000)$.

Then $Q$ lines,

1 s: add a new string s

2 a b c d: find how many strings satisfying the requirement above.

$\sum |s|+|a|+|b|+|c|+|d|\le 2000000$.

Output Format

For type $2$ query. Output the answer in one line.

樣例輸入

1
10
1 abcqaq
1 abcabcqaqqaq
2 ab bc qa aq
2 a c q q
1 abcabcqaqqwq
2 ab bc qa aq
2 a c q q
1 abcq
2 a c q q
2 a b c q

樣例輸出

1
2
1
3
3
1

解：因為題目沒有明確給出字串的個數與長度 所以這裡我們開一個長整型指標陣列避免越界

一開始逐個字元匹配T了 所以加了hash預處理

坑點 並不是文字串的總長度大於4個字串的總長度就可以 需要前兩個 和 後兩個字串的長度總和小於等於 文字串的一半

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <bitset>
#include <algorithm>
#include <climits>
using namespace std;
const int N = 1e6+10;
typedef long long LL;
char str[N], a[N], b[N], c[N], d[N];
const LL mod = 1e9+7;
const LL seed = 1313;
LL *hash1[N], pos[N];
int xlen[N];

int main()
{
    pos[0]=1;
    for(int i=1;i<N;i++) pos[i]=pos[i-1]*seed%mod;
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        int h=0;
        while(n--)
        {
            int xx;
            scanf("%d", &xx);
            if(xx==1)
            {
                scanf("%s",str+1);
                int l=strlen(str+1);
                xlen[h]=l;
                hash1[h]=new LL[l+3];
                hash1[h][0]=0;
                for(int i=1;i<=l;i++)
                    hash1[h][i]=(hash1[h][i-1]*seed%mod+(str[i]-'a'+1))%mod;
                h++;
            }
            else
            {
                scanf("%s %s %s %s",a+1,b+1,c+1,d+1);
                int l1=strlen(a+1),l2=strlen(b+1),l3=strlen(c+1),l4=strlen(d+1);
                int sum=l1+l2+l3+l4, cnt=0;
                LL ans1=0, ans2=0, ans3=0, ans4=0;
                for(int j=1; j<=l1; j++) ans1=(ans1*seed%mod+(a[j]-'a'+1))%mod;
                for(int j=1; j<=l2; j++) ans2=(ans2*seed%mod+(b[j]-'a'+1))%mod;
                for(int j=1; j<=l3; j++) ans3=(ans3*seed%mod+(c[j]-'a'+1))%mod;
                for(int j=1; j<=l4; j++) ans4=(ans4*seed%mod+(d[j]-'a'+1))%mod;

                for(int i=0;i<h;i++)
                {
                    if((xlen[i]&1)||l1+l2>xlen[i]/2||l3+l4>xlen[i]/2) continue;
                    int cx=xlen[i]/2;
                    LL x=hash1[i][l1];
                    if(x!=ans1) continue;
                    x=(hash1[i][cx]-hash1[i][cx-l2]*pos[l2]%mod+mod)%mod;
                    if(x!=ans2) continue;
                    x=(hash1[i][cx+l3]-hash1[i][cx]*pos[l3]%mod+mod)%mod;
                    if(x!=ans3) continue;
                    x=(hash1[i][2*cx]-hash1[i][2*cx-l4]*pos[l4]%mod+mod)%mod;
                    if(x!=ans4) continue;
                    cnt++;
                }
                printf("%d\n",cnt);
            }
        }
    }
    return 0;
}