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())
a = list(map(int,input().split()))
if(a == [i for i in range(1,n+1)]):
    print(0,0)
else:
    if(len(a)==2):
        print(1,2)
        exit()
    c = []
    for i in range(len(a)-1):
        if(a[i]==a[i+1]+1):
            c.append(i)
    if(len(c)!=0):
        c.append(c[-1]+1)
        l = c[0]+1
        r = c[-1]+1
        p = []
        p+=a[:l-1]
        p+=a[r-1:l-2:-1]
        p+=a[r:]
        if(p==[i for i in range(1,n+1)]):
            print(l,r)
        else:
            print(0,0)
    else:
        print(0,0)
