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.
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.
def solve(a,sol):
    f=0
    s=0
    for i in range(len(a)):
        if(a[i]!=i+1):
            f=i
            break
    
    for j in range(len(a)-1,-1,-1):
        if(a[j]!=j+1):
            s=j
            break
    
    x=a[f:s+1]
    x.reverse()
    if(a[:f]+x+a[s+1:]==sol):
        print(f+1,s+1)
    else:
        print(0,0)
    


n=int(input())
a=list(map(int,input().split()))
sol=[(i+1) for i in range(n)]
if(a==sol):
    print(0,0)
else:    
    solve(a,sol)