Codeforces Round #307 (Div. 2) B. ZgukistringZ

Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one.

GukiZ has strings a, b, and c. He wants to obtain string k by swapping some letters in a, so that k should contain as many non-overlapping substrings equal either to b or c as possible. Substring of string x is a string formed by consecutive segment of characters from x. Two substrings of string x overlap if there is position i in string x occupied by both of them.

GukiZ was disappointed because none of his students managed to solve the problem. Can you help them and find one of possible strings k?

Input

The first line contains string a, the second line contains string b, and the third line contains string c (1 ≤ |a|, |b|, |c| ≤ 105, where |s| denotes the length of string s).

All three strings consist only of lowercase English letters.

It is possible that b and c coincide.

Output

Find one of possible strings k, as described in the problem statement. If there are multiple possible answers, print any of them.

Examples

input

Copy

aaa
a
b

output

Copy

aaa

input

Copy

pozdravstaklenidodiri
niste
dobri

output

Copy

nisteaadddiiklooprrvz

input

Copy

abbbaaccca
ab
aca

output

Copy

ababacabcc

Note

In the third sample, this optimal solutions has three non-overlaping substrings equal to either b or c on positions 1 – 2 (ab), 3 – 4 (ab), 5 – 7 (aca). In this sample, there exist many other optimal solutions, one of them would be acaababbcc.题意：给出a b c三个字符数组，求出b c 是a中不相交字符的子串的最多个数

思路：枚举，先枚举出a中存在b的个数，再在a中减去b的子串的字符，然后再寻找c的个数。

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 100005;
const int M = 30;
char s[3][N];
int vis[3][M];
int ans, a, b;

int main()
{
    for (int i = 0; i < 3; i++) {
        scanf("%s", s[i]);
        int l = strlen(s[i]);
        for (int j = 0; j < l; j++) {
            vis[i][s[i][j] - 'a']++;
        }
    }
    ans = 0;
    for(int i=0; ;i++){
        int flag = N;
        for(int j = 0;j < 26; j++){
            if(vis[0][j]<vis[1][j]*i){
                flag = 0;
                break;
            }
        }
        if(!flag)
            break;
        int p = N;
        for(int j=0;j<26;j++){
            if(vis[2][j]){
                p=min(p,(vis[0][j]-vis[1][j]*i)/vis[2][j]);
            }
        }
        if(i+p>ans){
            ans=i+p;
            a=i;
            b=p;
        }
    }
    for(int i=0;i<a;i++)
        printf("%s",s[1]);
    for(int i=0;i<b;i++)
        printf("%s",s[2]);
    for(int i=0;i<26;i++){
        for(int j=0;j<vis[0][i]-vis[1][i]*a-vis[2][i]*b;j++)
            printf("%c",i+'a');
    }
    printf("\n");
    return 0;
}