Vasya collects coins: he has exactly one coin for every year from 1 to n. Naturally, Vasya keeps all the coins in his collection in the order in which they were released. Once Vasya's younger brother made a change — he took all the coins whose release year dated from l to r inclusively and put them in the reverse order. That is, he took a certain segment [l, r] and reversed it. At that the segment's endpoints did not coincide. For example, if n = 8, then initially Vasya's coins were kept in the order 1 2 3 4 5 6 7 8. If Vasya's younger brother chose the segment [2, 6], then after the reversal the coin order will change to 1 6 5 4 3 2 7 8. Vasya suspects that someone else could have spoilt the permutation after his brother. Help him to find that out. Check if the given permutation can be obtained from the permutation 1 2 ... n using exactly one segment reversal. If it is possible, find the segment itself.

Input

The first line contains an integer n (1 ≤ n ≤ 1000) which is the number of coins in Vasya's collection. The second line contains space-separated n integers which are the spoilt sequence of coins. It is guaranteed that the given sequence is a permutation, i.e. it contains only integers from 1 to n, and every number is used exactly 1 time.

Output

If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 ≤ l < r ≤ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.

Examples

Input

8
1 6 5 4 3 2 7 8


Output

2 6


Input

4
2 3 4 1


Output

0 0


Input

4
1 2 3 4


Output

0 0
import sys

######################################################
out = sys.stdout
_buf, _cur = ''.join(sys.stdin.readlines()).split(), 0
def rls(n = 1):
  global _cur
  _cur += n
  return _buf[_cur-n:_cur]
def rli(n = 1): return [int(x) for x in rls(n)]
def rlf(n = 1): return [float(x) for x in rls(n)]
def rs(): return rls()[0]
def ri(): return int(rs())
######################################################

n = ri()
a = rli(n)

def check (a, n):
  b = []
  for k in range(1, len(a)+1):
    if a[k-1] != k:
      b.append (k)
  if len(b) < 2:
    return '0 0'
  """
  for i in range(1, len(b)):
    if b[i-1] != b[i] - 1:
      return '0 0'
  """  
  i, j = b[0] - 1, b[-1] - 1
  c = a[:]
  while i < j:
    c[i], c[j] = c[j], c[i]
    i += 1
    j -= 1
  d = list(range(1, n+1))
  if c == d:
    return str(b[0]) + ' ' + str(b[-1])
  return '0 0'

out.write( check(a, n) + '\n')
