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.
def getSig(s):
  freq = {}
  for char in s:
    if char in freq:
      freq[char] += 1
    else:
      freq[char] = 1
  return freq

def getCount(x,y):
  temp = []
  for k in y:
    if k not in x:
      return 0
    cal = int(x[k]/y[k])
    if cal == 0:
      return 0
    temp.append(cal)
  return min(temp)

a = input()
b = input()
c = input()

sig_a = getSig(a)
sig_b = getSig(b)
sig_c = getSig(c)


cnt_b = getCount(sig_a,sig_b)
cnt_c = getCount(sig_a,sig_c)

# print(sig_a)
# print(sig_b)
# print(sig_c)

# print(cnt_b)
# print(cnt_c)

ans = []
if cnt_b > 0 or cnt_c > 0:
  if cnt_b > cnt_c:

    ans.append(b*cnt_b)
    
    for k in sig_b:
      sig_a[k] -= (sig_b[k]*cnt_b)

    cnt_c = getCount(sig_a,sig_c)
    ans.append(c*cnt_c)

    if cnt_c > 0:
      for k in sig_c:
        sig_a[k] -=  (sig_c[k]*cnt_c)

  else:

    ans.append(c*cnt_c)

    for k in sig_c:
      sig_a[k] -= (sig_c[k]*cnt_c)

    cnt_b = getCount(sig_a,sig_b)
    ans.append(b*cnt_b)

    if cnt_b > 0:
      for k in sig_b:
        sig_a[k] -= (sig_b[k]*cnt_b)

for k in sig_a:
  ans.append(k*sig_a[k])
print(''.join(ans))
