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.

Sample test(s)

input

aaa
a
b

output

aaa

input

pozdravstaklenidodiri
niste
dobri

output

nisteaadddiiklooprrvz

input

abbbaaccca
ab
aca

output

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.

这题因为字母可以随意调换，所以先数出a,b,c三个字符串所有字母表中的字母的个数，然后看最多能填充多少个b字符串minx,然后填充0~minx个字符串，再看在当前情况下能填充多少个c字符串，计算最大值。

#include<stdio.h>
#include<string.h>
char s1[100006],s2[100006],s3[100006];
int a[30],b[30],c[30],a1[30];
int main()
{
    int n,m,i,j,len1,len2,len3,minx1,minx2,maxx,minx,t1,t2;
    while(scanf("%s",s1)!=EOF)
    {
        scanf("%s%s",s2,s3);
        len1=strlen(s1);
        len2=strlen(s2);len3=strlen(s3);
        memset(a,0,sizeof(a));memset(b,0,sizeof(b));memset(c,0,sizeof(c));memset(a1,0,sizeof(a1));
        for(i=0;i<len1;i++){
            a[s1[i]-'a']++;a1[s1[i]-'a']++;
        }
        for(i=0;i<len2;i++){
            b[s2[i]-'a']++;
        }
        for(i=0;i<len3;i++){
            c[s3[i]-'a']++;
        }
        minx1=200006;
        for(i=0;i<=25;i++){
            if(b[i]==0)continue;
            if(minx1>(a[i]/b[i]))minx1=a[i]/b[i];
        }
        minx2=200006;
        for(i=0;i<=25;i++){
            if(c[i]==0)continue;
            if(minx2>(a[i]/c[i]))minx2=a[i]/c[i];
        }
        maxx=minx2;t1=0;t2=minx2;
        for(i=1;i<=minx1;i++){
            minx=200006;
            for(j=0;j<=25;j++){
                a[j]-=b[j];
            }
            for(j=0;j<=25;j++){
                if(c[j]==0)continue;
                if(minx>(a[j]/c[j]))minx=a[j]/c[j];
            }
            if(maxx<minx+i){
                maxx=minx+i;
                t1=i;t2=minx;
            }
        }
        for(i=1;i<=t1;i++)printf("%s",s2);
        for(i=1;i<=t2;i++)printf("%s",s3);
        for(i=0;i<=25;i++){
            a1[i]=a1[i]-t1*b[i]-t2*c[i];
            for(j=1;j<=a1[i];j++){
                printf("%c",'a'+i);
            }
        }
        printf("\n");
    }
    return 0;
}