import sys

k = int(sys.stdin.readline().strip())
for i in range(k):
  line = sys.stdin.readline().strip().split()
  alph = line[0]
  u = line[1]
  v = line[2]

  # first, see if either is a substring of the other (including equals):
  if u.find(v)>= 0 or v.find(u) >= 0:
    print(1)
    continue

  # We know that neither of u,v is a substring of the other.
  # If there are more than 2 letters in the alphabet, not regular.
  # [Suppose alphabet = {a,b,c}, suppose that c is neither the first
  # nor the last letter of v. Then it's not hard to show that, for
  # suffiently large n, u c^n u does not contain v.]
  if len(alph) > 2:
    print(0)
    continue

  # Two-letter alphabet. Make u the longer string:
  if len(u) < len(v):
    temp = u
    u = v
    v = temp

  ul = len(u)
  vl = len(v)

  # Since we know v is not a substring of u, v must contain include both
  # letters of the alphabet. [E.g., if v = a^n, then v is not contained in
  # u b u and we can "pump" strings of the form (ub)^n to get illegal words.]
  # Then it is sufficient to see whether v is contained in u x u
  # for each x in alph U {eps} [eps = empty string]. [Proof not hard?]
  # (If |u|==|v|, also do symmetric check)

  ok = True
  # check eps
  t = u[ul-vl+1:]+u[:vl-1]
  ok = t.find(v)>= 0
  #print("   empty1:",ok)

  if not ok and ul==vl: # u, v have same length, so symmetric check:
    t = v[1:]+v[ul-1]
    ok = t.find(u)>= 0
    #print("empty2",ok)

  for c in alph:
    if not ok: break
    t = u[ul-vl+1:]+str(c)+u[:vl-1]
    ok = t.find(v,0)>= 0
    #print("t1 = u[ul-vl+1:]+str(c)+u[:vl-1] =",u[ul-vl+1:],str(c),u[:vl-1])
    #print(ok)
    if not ok and ul==vl:
      t = v[1:]+str(c)+v[:ul-1]
      #print("t2 =  v[1:]+str(c)+v[:ul-1] =",v[1:],str(c),v[:ul-1])
      ok = t.find(u,0) >= 0
      #print(ok)
    if not ok: break
  if ok: print(1)
  else: print(0)