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
import sys
raw = sys.stdin.readlines()
x = 0
data = []
for x in raw:
    data.append(x.strip())

A = data[1].split()
for x in A:
    A[int(x)-1] = int(A[int(x)-1])

flag = 0
for z in range(int(data[0])-1):
    if A[z] != z+1:
        flag = 1
        start = z + 1
        break

if flag == 0:
    print '0 0'
else:
    turn = int(data[0])
    for i in range(turn):
        if A[turn - i - 1] != turn - i:
            end = turn - i
            break
    C = A[:]
    yop = start-1
    j = 1
    while yop < end:
        C[yop] = A[end-j]
        j += 1
        yop += 1
    for turn2 in range(len(C)):
        if C[turn2] != turn2 + 1:
            print '0 0'
            break
    else:
        print start, end
