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 res():
    n = int(input())
    v = [int(i) for i in input().split()]
    
    ans = sorted(v)
    
    if(v == ans):
        return "0 0"

    try:
        ini = 0
        temp = []
        
        while v[ini] < v[ini+1]:
            temp.append(v[ini])
            ini += 1
        
        fim = ini
        v2 = []
        while fim < n-1 and v[fim] > v[fim+1]:
            v2.append(v[fim])
            fim += 1
        v2.append(v[fim])
        
        ini += 1
        fim += 1
        
        temp += v2[::-1] + v[fim:]
        
        if temp == ans:
            return str(ini) + ' ' + str(fim)
        else:
            return "0 0"
        
    except:
        return "0 0"
        
def main():
    print(res())

main()
 				 			 	     	 	 	 	       	