Returning back to problem solving, Gildong is now studying about palindromes. He learned that a palindrome is a string that is the same as its reverse. For example, strings “pop”, “noon”, “x”, and “kkkkkk” are palindromes, while strings “moon”, “tv”, and “abab” are not. An empty string is also a palindrome.
Gildong loves this concept so much, so he wants to play with it. He has n
distinct strings of equal length m. He wants to discard some of the strings (possibly none or all) and reorder the remaining strings so that the concatenation becomes a palindrome. He also wants the palindrome to be as long as possible. Please help him find one.
Input
The first line contains two integers n
and m (1≤n≤100, 1≤m≤50) — the number of strings and the length of each string.
Next n lines contain a string of length m each, consisting of lowercase Latin letters only. All strings are distinct.
Output
In the first line, print the length of the longest palindrome string you made.
In the second line, print that palindrome. If there are multiple answers, print any one of them. If the palindrome is empty, print an empty line or don’t print this line at all.
Examples
Input
3 3
tab
one
bat
Output
6
proved
Input
4 2
oo
ox
xo
xx
Output
6
oxxxxo
Input
3 5
hello
codef
orces
Output
0
Input
9 4
abab
no
abcd
bcde
cdef
defg
wxyz
zyxw
ijji
Output
20
of bwxyzi jjizyxwbaba
Note
In the first example, “battab” is also a valid answer.
In the second example, there can be 4 different valid answers including the sample output. We are not going to provide any hints for what the others are.
In the third example, the empty string is the only valid palindrome string.
Process
In the contest, I didn’t solve this problem. I think too much.When I saw this code firstly, it’s so beautiful! Come on, i will do it.
Code
#include<bits/stdc++.h>
using namespace std;
int n, m;
string p, ans;
set<string> s;
int main()
{
cin >> n >> m;
for (int i = 0; i < n; i++)
{
string t, r;
cin >> t;
r = t;
s.insert(t);
reverse(r.begin(), r.end());
if (r == t)
p = t;
else if (s.count(r))
ans += r;
}
string t = ans;
reverse(ans.begin(), ans.end());
ans += p + t;
cout << ans.size() << endl << ans;
return 0;
}
