Codeforces Round #414 C.Naming Company（贪心）

C. Naming Company

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Oleg the client and Igor the analyst are good friends. However, sometimes they argue over little things. Recently, they started a new company, but they are having trouble finding a name for the company.

To settle this problem, they've decided to play a game. The company name will consist of n letters. Oleg and Igor each have a set of n letters (which might contain multiple copies of the same letter, the sets can be different). Initially, the company name is denoted by n question marks. Oleg and Igor takes turns to play the game, Oleg moves first. In each turn, a player can choose one of the letters c in his set and replace any of the question marks with c. Then, a copy of the letter c is removed from his set. The game ends when all the question marks has been replaced by some letter.

For example, suppose Oleg has the set of letters {i, o, i} and Igor has the set of letters {i, m, o}. One possible game is as follows :

Initially, the company name is ???.

Oleg replaces the second question mark with 'i'. The company name becomes ?i?. The set of letters Oleg have now is {i, o}.

Igor replaces the third question mark with 'o'. The company name becomes ?io. The set of letters Igor have now is {i, m}.

Finally, Oleg replaces the first question mark with 'o'. The company name becomes oio. The set of letters Oleg have now is {i}.

In the end, the company name is oio.

Oleg wants the company name to be as lexicographically small as possible while Igor wants the company name to be as lexicographically large as possible. What will be the company name if Oleg and Igor always play optimally?

A string s = s₁s₂...s_m is called lexicographically smaller than a string t = t₁t₂...t_m (where s ≠ t) if s_i < t_i where i is the smallest index such that s_i ≠ t_i. (so s_j = t_j for all j < i)

Input

The first line of input contains a string s of length n (1 ≤ n ≤ 3·10⁵). All characters of the string are lowercase English letters. This string denotes the set of letters Oleg has initially.

The second line of input contains a string t of length n. All characters of the string are lowercase English letters. This string denotes the set of letters Igor has initially.

Output

The output should contain a string of n lowercase English letters, denoting the company name if Oleg and Igor plays optimally.

Examples

Input

tinkoff
zscoder

Output

fzfsirk

Input

xxxxxx
xxxxxx

Output

xxxxxx

Input

ioi
imo

Output

ioi

Note

One way to play optimally in the first sample is as follows :

• Initially, the company name is ???????.
• Oleg replaces the first question mark with 'f'. The company name becomes f??????.
• Igor replaces the second question mark with 'z'. The company name becomes fz?????.
• Oleg replaces the third question mark with 'f'. The company name becomes fzf????.
• Igor replaces the fourth question mark with 's'. The company name becomes fzfs???.
• Oleg replaces the fifth question mark with 'i'. The company name becomes fzfsi??.
• Igor replaces the sixth question mark with 'r'. The company name becomes fzfsir?.
• Oleg replaces the seventh question mark with 'k'. The company name becomes fzfsirk.

For the second sample, no matter how they play, the company name will always be xxxxxx.

分析：

由题意，第一个人要构造最小的字符串，而第二个人要构造最大的字符串。

当n为奇数时 集合一中要用到n/2+1个字母 ,n为偶数时用到n/2个字母，那么我们可以只保留字典序较小的字母

集合二则只保留字典序较大的字母轮到集合一取数的时候

当集合一中的最小的字母小于集合二中最大的字母时，则将该字母放在字符串的下一个位置，否则，将集合一中最大的字母放在字符串从后往前数的下一个位置。轮到集合二取数的时候

当集合二中最大的字母大于集合一中最小的字母时，将该字母放在字符串的下一个位置，否则，将集合二中最小的字母放在字符串从后往前数的下一个位置。

这样就能构造出符合的字符串。

代码如下：

#include <cstdio>
#include <algorithm>
#include <iostream>
#include<cstring>
using namespace std;
char a[300010];
char b[300010];
char res[300010];
int cmp(int x,int y)
{
  return x>y;
}
int main()
{
	int na,nb,cnt,pa,qa,pb,qb,l,r,L;
	while(scanf("%s%s",a,b)!=EOF)
	{
		cnt=0;
		na=strlen(a);
		L=na;
		nb=strlen(b);
		sort(a,a+na);
		sort(b,b+nb,cmp);
		if(na%2==1)
		na=na/2+1;
		else
		na=na/2;
		nb=L-na;
		pa=0;
		qa=na-1;
		pb=0;
		qb=nb-1;
		l=0;
		r=L-1;
		for(int i=0;i<L;i++)
		{
		   if(i%2==0)
		   {
		   	 if(a[pa]>=b[pb])
		   	  res[r--]=a[qa--];
		   	  else
		   	  res[l++]=a[pa++];
		   }
		   else
		   {
		   	  if(a[pa]>=b[pb])
		   	  	res[r--]=b[qb--];
		   	  else
		   	  res[l++]=b[pb++];
		   }
		}
		res[L]='\0';
		printf("%s",res);
		printf("\n");
	}
	return 0;
}