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()]
f=0
if l==sorted(l):
    print('0 0')
    exit()
if len(set(l))!=n:
    print('0 0')
    exit()
f=0
for i in range(0,n):
    if l[i]!=(i+1):
        f=1 
        indx=l.index(i+1)
        curr=i 
        arr=l[i:indx+1]
        arr=arr[::-1]
        def check(arr):
            n=len(arr)
           # print(arr)
            for i in range(1,n):
                if arr[i]!=arr[i-1]+1:
                    return 0
            return 1
        if check(arr):
            print(curr+1,indx+1)
        else:
            print('0 0')
        exit()
print('0 0')