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(" ")))
fl1=False;fl2=False;b=a[0];t1=0;t2=n;d=[];fl3=True
for i in range(len(a)):
    if (a[i]<b):
        t1=i
        fl1=True
        break
    b=a[i]
if (fl1):
    b=a[t1]
    for i in range(t1+1,n):
        if (a[i]>b):
            t2=i
            fl2=True
            break
        b=a[i]
if (fl1):
    b=0
    for i in range(t1-1):
        d.append(a[i])
    for i in range(t2-1,t1-2,-1):
        d.append(a[i])
    for i in range(t2,n):
        d.append(a[i])
    for i in range(n):
        if (d[i]<b):
            fl3=False
            break
        b=d[i]
    if (fl3):
        print(t1,t2)
    else:
        print("0 0")
else:
    print("0 0")