Naming Company CodeForces

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 The game ends when all the question marks has been replaced by some letter.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

Examples

Input

tinkoff
zscoder

Output

fzfsirk

Input

xxxxxx
xxxxxx

Output

xxxxxx

Input

ioi
imo

Output

ioi

题意：甲乙两人各持有一个长度均为n的字符串，轮着向一个新的长也为n的字符串里放字符，甲先行。
甲每一步都试图让字符串按字典序最小化，乙每一步都试图让字符串按字典序最大化。问最后这新字符串是什么。

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define  pi acos(-1.0)
#define  eps 1e-6
#define  fi first
#define  se second
#define  lson l,m,rt<<1
#define  rson m+1,r,rt<<1|1
#define  bug         printf("******
")
#define  mem(a,b)    memset(a,b,sizeof(a))
#define  fuck(x)     cout<<"["<<x<<"]"<<endl
#define  f(a)        a*a
#define  sf(n)       scanf("%d", &n)
#define  sff(a,b)    scanf("%d %d", &a, &b)
#define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  pf          printf
#define  FRE(i,a,b)  for(i = a; i <= b; i++)
#define  FREE(i,a,b) for(i = a; i >= b; i--)
#define  FRL(i,a,b)  for(i = a; i < b; i++)
#define  FRLL(i,a,b) for(i = a; i > b; i--)
#define  FIN         freopen("DATA.txt","r",stdin)
#define  gcd(a,b)    __gcd(a,b)
#define  lowbit(x)   x&-x
#pragma  comment (linker,"/STACK:102400000,102400000")
using namespace std;
typedef long long  LL;
typedef unsigned long long ULL;
const int INF = 0x7fffffff;
const int mod = 1e9 + 7;
const int maxn = 4e5 + 10;
int n, k, a[maxn], ans[maxn];
char s1[maxn], s2[maxn], s3[maxn];
int cmp(char x, char y) {
    return x > y;
}
int main() {
    scanf("%s%s", s1, s2);
    int n = strlen(s1), k = 0;
    sort(s1, s1 + n);
    sort(s2, s2 + n, cmp);
    int i = 0, j = 0, ed1 = (n - 1) / 2, ed2 = (n - 1) / 2, L = 0, R = n - 1;
    if (n & 1) ed2--;
    while(k < n) {
        if (k % 2 == 0) {
            if (s1[i] >= s2[j]) s3[R--] = s1[ed1--];
            else s3[L++] = s1[i++];
        } else {
            if (s2[j] <= s1[i]) s3[R--] = s2[ed2--];
            else s3[L++] = s2[j++];
        }
        k++;
    }
    s3[n] = '';
    printf("%s
", s3);
    return 0;
}