CF508D：Tanya and Password（欧拉通路 & 输出路径）

D. Tanya and Password

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.

Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.

Input

The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got.

Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.

Output

If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".

If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.

Examples

input

5acaabaabacabbac

output

YESabacaba

input

4abcbCbcb1b13

output

NO

input

7aaaaaaaaaaaaaaaaaaaaa

output

YESaaaaaaaaa

题意：给N个长度为3的字符串，问能否组成总长度为N+2的字符串，如abcde=abc+bcd+cde，能就输出给字符串。

思路：每个字串后两位要跟另一个串的前两位一样才能接起来，那么将每个串的前两位和后两位拆开Hash后建边，跑一遍欧拉通路就行。

//referencr：UniqueColor# include <bits/stdc++.h># define pb push_backusing namespace std;const int maxn = 2e5+30;vector<int>edge[maxn];int cnt[maxn], in[maxn], out[maxn];string ans;void dfs(int x){    while(cnt[x] < edge[x].size())        dfs(edge[x][cnt[x]++]);    ans += (char)(x%256);}int main(){    int n, u, v;    char s[5];    scanf("%d",&n);    for(int i=0; i<n; ++i)    {        scanf("%s",s);        u = s[0]*256+s[1];        v = s[1]*256+s[2];        edge[u].pb(v);        ++out[u];        ++in[v];    }    int _s=0, _e=0, st=u;//有可能起点和终点的入出度差均为0。    for(int i=0; i<maxn; ++i)    {        int sub = in[i] - out[i];        if(sub == -1) ++_s, st=i;        else if(sub == 1) ++_e;        else if(sub != 0)return 0*puts("NO");    }    if(_s>1 || _e>1 || _s + _e == 1) return 0*puts("NO");    dfs(st);    ans += (char)(st/256);    if(ans.size() != n+2) puts("NO");    else    {        puts("YES");        reverse(ans.begin(), ans.end());        cout << ans << endl;    }    return 0;}