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())
m=input().split()
a=[]
s=0
s1=0
k=0
h=0
l=""
c=""
for i in m:
	a.append(int(i))
for i in range (n-1):
	if a[i]  > a[i+1] :
		s+=1
	if a[i] < a[i+1]:
		s1+=1
if s >= n/2:
	for i in range (n-1):
		if a[i]-1!= a[i+1]:
				k+=1
				if k ==2:
					l+=str(a[i])
					break
				else:
					l+=str(a[i+1])
					l+=' '
	if len(l) <3:
		print('0 0')
	else:
		if int(l[0]) > int(l[2]):
			print(l[2],l[0])
		else:
			print(l)
		
else:
	for i in range (n-1):
		if a[i] +1 != a[i+1]:
			k+=1
			if k == 2:
				l+=str(a[i])
				break
			else:
				l+=str(a[i+1])
				l+=' '
	if len(l) <3:
		print('0 0')
	else:
		if int(l[0]) > int(l[2]):
			print(l[2],l[0])
		else:
			print(l)