Naming Company CodeForces - 794C （博弈，构造）

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 The game ends when all the question marks has been replaced by some letter.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...sm is called lexicographically smaller than a string t = t1t2...tm (where s ≠ t) if si < ti where i is the smallest index such that si ≠ ti. (so sj = tj 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.

题意：

给你两个等长度的字符串s1,s2.，s1属于先手拥有，s2属于后手拥有

现在要新凑一个长度相等字符串，由先手后手交叉拿出一个字符串选择放在指定的位置。

先手想让新字符串字典序更小，后手想让新字符串的字典序更大。二者都绝顶聪明，请输出最终答案。

思路：

很容易产生一个错误的思路：

即s1从小到大排序，s2从大到小排序，二人从前到后依次选出。

而当s1中最小的字符比s2最大的字符大的时候，这样处理就会出现错误。

并且我们容易知道先手一定拿出s1的排序后的前半部分，s2也是。

那么我们把一定拿出的部分分别代替掉s1和s2串。

当s1中最小的字符比s2中最大的字符小的时候，在新字符串中从前往后填字符，

否则即s1中最小的字符比s2中最大的字符大的时候，

我们拿出s1中最大的字符，填在新字符串能填的最后的位置，

同时后手为了让字典序更大，应该把s2中最小的字符填在新字符串能填的最后的位置。

依次得出的新字符串就是答案，

细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;}
inline void getInt(int* p);
const int maxn = 1000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
string s1,s2;
int len;
char ans[maxn];
int main()
{
	//freopen("D:\\code\\text\\input.txt","r",stdin);
	//freopen("D:\\code\\text\\output.txt","w",stdout);
	gbtb;
	cin>>s1>>s2;
	len=s1.length();
	int l1=0;
	int l2=0;
	int r1=(len+1)/2-1;
	int r2=len-r1-2;
	sort(ALL(s1));
	sort(ALL(s2),greater<char>());
	int l=0;
	int r=len-1;
//	repd(i,0,r1)
//	{
//	    cout<<s1[i];
//	}
//	cout<<endl;
//
//	repd(i,0,r2)
//	{
//	    cout<<s2[i];
//	}
//	cout<<endl;

	repd(i,1,len)
	{
		if(s1[l1]>=s2[l2])
		{
			if(i&1)
			{
				ans[r--]=s1[r1--];
			}else
			{
				ans[r--]=s2[r2--];
			}
		}else
		{
			if(i&1)
			{
				ans[l++]=s1[l1++];
			}else
			{
				ans[l++]=s2[l2++];
			}
		}
	}
	ans[len]='\0';
	cout<<ans<<endl;
	return 0;
}

inline void getInt(int* p) {
	char ch;
	do {
		ch = getchar();
	} while (ch == ' ' || ch == '\n');
	if (ch == '-') {
		*p = -(getchar() - '0');
		while ((ch = getchar()) >= '0' && ch <= '9') {
			*p = *p * 10 - ch + '0';
		}
	}
	else {
		*p = ch - '0';
		while ((ch = getchar()) >= '0' && ch <= '9') {
			*p = *p * 10 + ch - '0';
		}
	}
}