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())
a = list(map(int, input().split()))

l = 0
r = 0
i = 0
while i < len(a):
    if i + 1 != a[i]:
        if i + 1 != a[a[i] - 1]:
            break
        if 0 != l and 0 != r:
            l = 0
            r = 0
            break
        sb = False
        for j in range(i, a[i] - 2):
            if a[j] != a[j + 1] + 1:
                sb = True
                break
        if sb:
            break

        l = i + 1
        r = a[i]
        i = r

    i += 1

if 0 != l and 0 != r:
    print(l, r)
else:
    print(0, 0)