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:
def main():
    num = int(input())
    data = [int(x) for x in input().split()]
    left = 0
    right = 0
    sign = True
    i = 0
    j = 0

        
    if len(data)<2:
        sign = False
    #find left
    for i in range(len(data)):
        if data[i] != i+1:
            break
    left = i+1
    
    #check left
    if left>num:
        sign = False   
    #find right
    for j in range(num-1, i,-1):
        if data[j] != j+1:
            break           
    right = j+1
    
    #check left
    if right<1:
        sign = False
    if left > right:
        sign = False
    #check data from left to right
    for i in range(left-1, right-1):
       
        if data[i]!= data[i+1]+1:
            sign = False
            break

        
   
    if sign:
        print(left,right)
    else:
        print("0 0\n")

main()