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.
def ascending_until(l):
    for k, e in enumerate(l):
        if k + l[0] != e:
            if k == 1:
                return 0
            else:
                return k

    return len(l)


def descending_until(l):
    for k, e in enumerate(reversed(l)):
        if l[-1] - k != e:
            if k == 1:
                return len(l)
            else:
                return len(l) - k

    return 0


if __name__ == '__main__':
    n = int(input())
    order = list(map(int, input().split()))

    i = ascending_until(order)
    if i == len(order):
        print(0, 0)
    else:
        j = descending_until(order)
        order[i:j] = list(reversed(order[i:j]))
        if ascending_until(order) == len(order):
            print(order[i], order[j - 1])
        else:
            print(0, 0)
