D - Naming Company CodeForces - 794C贪心

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 cin 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 The game ends when all the question marks has been replaced by some letter.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_jfor 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.

Example

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.

题意：Oleg和Igor都有一串长度为n的字符串，现在有一串由n个"?"组成的字符串，两人用自己的字符串中的字符填补这个？组成的字符串 ，Oleg想让这个字符串尽量小，Igor想让字符串尽量大，问最后的字符串。

思路：第一个字符从小到大排序，第二个从大到小排序，比较当前要放的字符，如果自己的字符小，或者自己的字符大，就把自己最左的字符放在最左，反之，把自己最右字符放在最右。

#include<bits/stdc++.h>
using namespace std;
#define maxn 300005
int n;
char str1[maxn],str2[maxn];
bool cmp(char a,char b)
{
    return a>b;
}
int main()
{
    while(~scanf("%s%s",str1,str2)){
            n=strlen(str1);
            char str[maxn];
        sort(str1,str1+n);
        sort(str2,str2+n,cmp);
        for(int i=0,ls=0,lt=0,rs=(n-1)/2,l=0,rt=(n-2)/2,r=n-1;i<n;i++){
            if(i&1){//放t
                if(str2[lt]>str1[ls]) str[l++]=str2[lt++];
                else str[r--]=str2[rt--];
            }
            else{
                if(str1[ls]<str2[lt]) str[l++]=str1[ls++];
                else str[r--]=str1[rs--];
            }
        }
        int i=0;
        while(i<n) putchar(str[i++]);
        putchar('\n');
    }
}