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*?
The first line contains string *a*, the second line contains string *b*, and the third line contains string *c* (1 ≤ |*a*|, |*b*|, |*c*| ≤ 10<sup class="upper-index">5</sup>, 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.
a=input()
b=input()
c=input()

k=a
lena = len(a)
lenb = len(b)
lenc = len(c)
flag =0
count1 = 0
count2 = 0
k1 = a
k2 = a

l=lena
while(l>=lenb):
    i=0
    while i<lenb:

        x = k1.find(b[i],0,l-i)

        if x>=0:

            k1 = k1[:x]+k1[x+1:]+b[i]
        else:
            flag = 1
            break
        i = i+1

    if(flag==1):
        k1=k1[lena-i:]+k1[:lena-i]
        break
    l=l-lenb

count1b = k1.count(b)

flag = 0
while(l>=lenc):
    i=0
    while i<lenc:
        x= k1.find(c[i],0,l-i)
        if x>=0:
            k1= k1[:x]+k1[x+1:]+c[i]
        else:
            flag = 1
            break
        i= i+1

    if flag == 1:
        k1=k1[lena-i:]+k1[:lena-i]
        break
    l = l-lenc

count1c = k1.count(c, 0, lena-count1b*lenb)

flag =0
l=lena
count1 = count1b+count1c
while(l>=lenc):
    i=0
    while i<lenc:

        x = k2.find(c[i],0,l-i)

        if x>=0:

            k2 = k2[:x]+k2[x+1:]+c[i]
        else:
            flag = 1
            break
        i = i+1

    if(flag==1):
        k2=k2[lena-i:]+k2[:lena-i]
        break
    l=l-lenc

count2c = k2.count(c)
flag = 0
if count2c == count1c:
    count2 = count1
else:
    while(l>=lenb):
        i=0
        while i<lenb:
            x= k2.find(b[i],0,l-i)
            if x>=0:
                k2= k2[:x]+k2[x+1:]+b[i]
            else:
                flag = 1
                break
            i= i+1

        if flag == 1:
            k2=k2[lena-i:]+k2[:lena-i]
            break
        l = l-lenb

    count2b = k2.count(b, 0, lena-count2c*lenc)
    count2 = count2b+count2c


if count1>=count2:
    print(k1)
else:
    print(k2)

    
