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.
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.
import sys

n_nums = int(sys.stdin.readline())
values = [int(x) for x in sys.stdin.readline().split()]

if n_nums == 1:
    print(0, 0)
    sys.exit(0)
    
if list(reversed(values)) == sorted(values):
    print(1, len(values))
    sys.exit(0)

start, end = False, False
s, e = 0, 0

for i in range(1, len(values)):
    if values[i] - values[i-1] < 0:
        if not start:
            start = True
            s = i
    if values[i] - values[i-1] > 0:
        if start and not end:
            end = True
            e = i
            break

in_tune = values[:s-1] + list(reversed(values[s-1:e])) + values[e:]
if in_tune == sorted(values):
    print(s, e)
else:
    print(0, 0)
