Problem: 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.

Input

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.

Output

If it is impossible to obtain the given permutation from the original one in exactly one action, print 0 0. Otherwise, print two numbers l r (1 ≤ l < r ≤ n) which are the endpoints of the segment that needs to be reversed to obtain from permutation 1 2 ... n the given one.

Examples

Input

8
1 6 5 4 3 2 7 8


Output

2 6


Input

4
2 3 4 1


Output

0 0


Input

4
1 2 3 4


Output

0 0

Solution (in PYTHON3):

Solution:
n = int(input())

L = [int(i) for i in input().split(" ")]

def main(n,L):
    since = -1
    end = -1
    if n == 1:
        return "0 0"
    
    if n == 2:
    	return "0 0" if L[0]<L[1] else "1 2"
    
    if L[1] == L[0]+1:
        order = 1
    elif L[1] == L[0]-1:
        order = -1
        since = 0
    elif L[2] == L[1]-1:
        order = -1
        since = 1
    else:
        return "0 0"

    for i in range(2,n):
        if L[i] != L[i-1] + order:
            if order == 1:
                if since == -1:
                    order = -1
                    since = i
                else:
                    return "0 0"
            else:
                end = i-1
                order = 1
        if end < since and end != -1:
            print(i)

    if end == -1 and since != -1:
        end = n-1

    since+=1
    end+=1
    if since == end:
        return "0 0"
    return "{} {}".format(since,end)
print(main(n,L))