Rohit collects coins: he has exactly one coin for every year from 1 to n. Naturally, Rohit keeps all the coins in his collection in the order in which they were released. Once Rohit'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 Rohit's coins were kept in the order 1 2 3 4 5 6 7 8. If Rohit'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. Rohit 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 which is the number of coins in Rohit'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.
-----Constraints-----
- $1 \leq N \leq 1000$
- $1 \leq A[N] \leq 10^9$
-----Sample Input:-----
8
1 6 5 4 3 2 7 8
-----Sample Output:-----
2 6
from sys import stdin,stdout,exit
from heapq import *
try:
    n=(int(stdin.readline()))
    a=list(map(int,stdin.readline().split()))
    s=0
    l=0
    r=n
    left=right=False
    for i in range(n):
        if s+1==a[i]:
            s=a[i]
        else:
            l=i
            s=a[l]
            break
    for i in range(l+1,n):
        if s-1==a[i]:
            s=a[i]
        else:
            s=a[i]
            r=i
            break
    arr = a[:l]+(a[l:r])[::-1]+a[r:]
    s=0
    ans=True
    for i in arr:
        if s+1==i:
            s=i
        else:
            ans=False
            break
    if ans and l+1!=r:
        print(l+1,r)
    else:
        print(0,0)
except:
    print(0,0)

from sys import stdin,stdout,exit
from heapq import *
try:
    n=(int(stdin.readline()))
    a=list(map(int,stdin.readline().split()))
    s=0
    l=0
    r=n
    left=right=False
    for i in range(n):
        if s+1==a[i]:
            s=a[i]
        else:
            l=i
            s=a[l]
            break
    for i in range(l+1,n):
        if s-1==a[i]:
            s=a[i]
        else:
            s=a[i]
            r=i
            break
    arr = a[:l]+(a[l:r])[::-1]+a[r:]
    s=0
    ans=True
    for i in arr:
        if s+1==i:
            s=i
        else:
            ans=False
            break
    if ans and l+1!=r:
        print(l+1,r)
    else:
        print(0,0)
except:
    print(0,0)
