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=input("")
n=int(n)
t=input("")
t=t.split(' ')
SP=True
Fin=False
for i in range(n):
    if int(t[i])!=i+1:
        k,r=i,int(t[i])
        SP=True
        break
    else:
        SP=False
if SP==True:    
    for i in range(0,n-k):
        if int(t[k+i])!=r-i:
            a,l=k+i,r-i
            Fin=False
            break
        else:
            Fin=True
    if Fin==False:
        for i in range(0,n-a):
            if int(t[a+i])!=r+i+1:
                SP=False
                break
        if (SP):
            print( str(l+1)+" "+str(r))
        else:
            print("0 0")
    else:
        print( str(k+1)+" "+str(r))
else:
    print("0 0")