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()))
p = []
k = True
for i in range(len(a)):
    if a[i] != i + 1:
        k = False
    if not k:
        p.append(a[i] + 1)
f = True
p1 = 0
p2 = 0
s = False
while len(p) != 0:
    q = p[0]
    for i in range(len(p)):
        if p[i] != q:
            break
        q -= 1
        if i == len(p) - 1:
            f = False
            p1 = p[i]
            p2 = p[0]
            s = True
    if s:
        break
    p.pop()
if not f and p1 != p2:
    print(p1 - 1, p2 - 1)
else:
    print(0, 0)
