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.
n = int(input())

L = [int(i) for i in input().split(" ")]

def main(n,L):
    prev = 0
    since = -1
    end = -1

    if L[1] == L[0]+1:
        order = 1
    elif L[1] == L[0]-1:
        order = -1
        since = 0
    else:
        if L[2] == L[1]+1:
            order = 1
            since = 0
            end = 0
        elif L[2] == L[1]-1:
            order = -1
            since = 1
        else:
            return "0 0"
    prev = L[1]

    for i in range(2,n):
        if L[i] != L[i-1] + order:
            if order == 1:
                if since != -1:
                    order = -1
                    since = i
                else:
                    return "0 0"
            if order == -1:
                end = i-1
                order = 1

    if end == -1 and since != -1:
        end = n-1

    since+=1
    end+=1
    return "{} {}".format(since,end)
print(main(n,L))
