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())
liste=input()
liste=liste.split(' ')
p=[]
for i in range(n):
    p.append(int(liste[i]))

correct=[]
perm=[]

for i in range(n):
    if p[i]!=i+1:
        perm.append(p[i])
    elif i>0 and p[i]==i+1 and p[i]==p[i-1]-1:
        perm.append(p[i])
    else:
        correct.append(p[i])
if perm:
    max=perm[0]
    min=perm[-1]
    m=len(perm)
    liste=[0]*m
    booleen=False
    for i in range(m):
        liste[i]=perm[m-1-i]
    if liste==list(range(min,max+1)):
        print(min,max)
    else:
        print('0 0')
else:
    print('0 0')