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())

inpList = [int(a) for a in input().split()]

normalList = [a for a in range(1,n+1)]

flag = False
flag2 = 0

i = 0

i1 = 0
i2 = 0

if inpList[0] != min(inpList) and inpList[-1] != max(inpList):
    print(0, 0)
    exit()

while i < n:
    if inpList[i] != normalList[i] and flag == False:
        flag = True
        flag2 += 1
        i1 = i+1
    elif flag == True and inpList[i] == normalList[i]:
        if flag2 == 2:
            i2 = i
            flag = False
        else:
            flag2 += 1
    i += 1
print(i1, i2)