E. Compress Words

Problem

Amugae has a sentence consisting of $n$ words. He want to compress this sentence into one word. Amugae doesn't like repetitions, so when he merges two words into one word, he removes the longest prefix of the second word that coincides with a suffix of the first word. For example, he merges "sample" and "please" into "samplease".

Amugae will merge his sentence left to right (i.e. first merge the first two words, then merge the result with the third word and so on). Write a program that prints the compressed word after the merging process ends.

Input

The first line contains an integer $n (1 \leq n \leq 10^5)$, the number of the words in Amugae's sentence.

The second line contains $n$ words separated by single space. Each words is non-empty and consists of uppercase and lowercase English letters and digits $('A', 'B', ..., 'Z', 'a', 'b', ..., 'z', '0', '1', ..., '9')$. The total length of the words does not exceed $10^6$.

Output

In the only line output the compressed word after the merging process ends as described in the problem.

Examples

input

5
I want to order pizza

output

Iwantorderpizza

input

5
sample please ease in out

output

sampleaseinout

idea

We want to connect to the first two words, and such qwer asdf these two, there is no overlapping portion, directly becomes qwerasdf; qwer erty duplicate and ER, becomes qweryt.

Two words done, then to consider more than one word asdf, qwer, erty, the result is asdfqwerty;

It is clear that only neighboring words have an impact on each other, then this question put big problem --n words put together - converted into a small problem - connected two by two

For short word directly matching algorithm BF violence is possible, but considering this question the data range, apparently to use KMP (in fact, a string hash is also OK but I will not)

KMP calculation time can be optimized, you need only compare two strings to the short length of string.

For example, I have qwertyuiopasdfghjklzxcv and ZXCVBNM, length 7 and 23, respectively, so long strings directly after seven and a short string;

Another example I have 90qwer and qwertyuiopasdfghjklzxcv then I just long strings of the first six and 90qwer.

Official explanations

Denote the words from left to right as $W_1,W_2,W_3,⋯,W_n$.

If we define string $F(k)$ as the result of merging as described in the problem $k$ times, we can get $F(k+1)$ by the following process:

If length of $F(k)$ $>$ length of $W_{k+1}$
Assume the length of $F(K)$ is $x$, and the length of $W_{k+1}$ is $y$. Construct the string $c=W_{k+1}+F(k) [ x−y...x ] $ ( * $s[x..y]$ for string $s$ is the substring from index $x$ to $y$)
Get the KMP failure function from string c.
We can get maximum overlapped length of $W_{k+1}$'s prefix and $F(k)$'s suffix from this function. Suppose the last element of the failure function smaller than the length of $W_{k+1}$ is $z$. Then the longest overlapped length of $F(k)$'s suffix and $W_{k+1}$'s prefix is $min(z,y)$. Let $L=min(z,y)$.
Then, $F(k+1)=F(k)+W_{k+1}[L+1...y]$
Otherwise
Construct $c$ as $W_{k+1}[1...x]+F(k)$. We can get $F(k+1)$ from the same process described in 1.
In this process, we can get $F(k+1)$ from $F(k)$ in time complexity $O(len(W_{k+1}))$. So, we can get $F(N)$ (the answer of this problem) in \ (H (referred to as (W_1) + only (W_2) + ⋯ + only (W_N)) \) .

The Code Of My Program

#include<cstdio>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<functional>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<numeric>
#include<cctype>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<map>
using namespace std;
 
const int N = 10000002;
int nxt[N];
char S[N], T[N],str1[N],str2[N];
int slen, tlen;
 
void getNext()
{
    int j, k;
    j = 0;
    k = -1;
    nxt[0] = -1;
    while(j < tlen)
        if(k == -1 || T[j] == T[k])
        {
            nxt[++j] = ++k;
            if (T[j] != T[k]) //KMP优化,加速得到next数组
                nxt[j] = k;
        }
        else
            k = nxt[k];
}
/*
返回模式串T在主串S中首次出现的位置
返回的位置是从0开始的
*/
int KMP_Index()
{
    int i = max(slen-tlen,0), j = 0;
    getNext();
    while(i < slen && j < tlen)
    {
        if(j == -1 || S[i] == T[j])
        {
            i++;
            j++;
        }
        else
            j = nxt[j];
    }
    return j;//T串前缀在主串S可以匹配的最远的可匹配的位置
}


int main()
{
 
    int n;
    scanf("%d",&n);
    scanf("%s",S);
    slen=strlen(S);
    for(int i=1; i<n; i++)
    {
        scanf("%s",T);
        tlen = strlen(T);
        int pos=KMP_Index();
        //printf(" pos =  %d  \n",pos);
        //cout<<pos<<endl;
        for(int j=pos;j<tlen;j++)
        {
            S[slen]=T[j];
            slen++;
        }
    }
    printf("%s",S);

return 0;
}

CodeForce Round 578 Div 2 E. Compress Words