Problem:
Score : 200 points

Problem Statement

Iroha has a sequence of N strings S_1, S_2, ..., S_N. The length of each string is L.

She will concatenate all of the strings in some order, to produce a long string.

Among all strings that she can produce in this way, find the lexicographically smallest one.

Here, a string s=s_1s_2s_3...s_n is lexicographically smaller than another string t=t_1t_2t_3...t_m if and only if one of the following holds:

There exists an index i(1≦i≦min(n,m)), such that s_j = t_j for all indices j(1≦j<i), and s_i<t_i.

s_i = t_i for all integers i(1≦i≦min(n,m)), and n<m.

Constraints

1 ≦ N, L ≦ 100

For each i, the length of S_i equals L.

For each i, S_i consists of lowercase letters.

Input

The input is given from Standard Input in the following format:

N L
S_1
S_2
:
S_N

Output

Print the lexicographically smallest string that Iroha can produce.

Sample Input 1

3 3
dxx
axx
cxx

Sample Output 1

axxcxxdxx

The following order should be used: axx, cxx, dxx.

Solution in JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)


let _inputLines;
let _lineNumber = 0;
let inputReader = _inputReader ();

function _main() {
	
	_inputLines = _inputData.trim().split('\n').map((string) => {
		return string.trim();
	});;
	
	let [a, b] = inputReader.readArray();
	//console.log(typeof c);
	a = parseInt(a);
	b = parseInt(b);
	//c = parseInt(c);
	let arr = [];
	for(let i=0;i<a;i++){
		let str = inputReader.readLine().split('').sort().join('');
		arr.push(str);
	} 
	arr.sort();
	let res='';
	for (let array of arr){
		res+=array;
	} 
	console.log(res);
	
	
	// let input = inputReader.readLine().split(' ');
	// let a = input[0];
	// let b = input[1];
	// //let b = inputReader.readLine().split('');
	// //console.log(a,b);
	
	// let res = '';
	// for(let i=0;i<size;i++){
	//     res +=a[i];
	//     res+=b[i];
	// }
	// console.log(res);
	
	//console.log(noOfTestCases,typeof noOfTestCases);
	// while(noOfTestCases--){
	//     let [a,b] = inputReader.readNumberArray();
	//     let count = 1;
	//     if(a<=2){
	//         console.log(1);
	//         continue;
	//     }
	//     let floor;
	//     if((a-2)%b===0)
	//         floor = (a-2)/b +1;
	//     else
	//          floor = Math.ceil((a-2)/b)+1;
	//     console.log(floor);
	
	// }

}

var _inputData = '';
function cacheInput(data) {
	_inputData += data;
}

process.stdin.resume();
process.stdin.setEncoding('utf8');
process.stdin.on('data', cacheInput).on('end', _main);

function _inputReader () {
	function readArray() {
		return _inputLines[_lineNumber++].split(' ');
	}
		
	function readLine(){
		return _inputLines[_lineNumber++];
	}
		
	function readNumberArray(){
		return _inputLines[_lineNumber++].split(' ').map(val => Number(val));
	}
		
	
	return {
		readArray,
		readLine,
		readNumberArray,
	}
}