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
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class R052Div2B {

	public static void main(String[] args) {
		FastScanner in=new FastScanner();
		int n=in.nextInt();
		int[] a=new int[n];
		int[] c=new int[n];
		for(int i=0;i<n;i++){
			a[i]=in.nextInt();
			c[i]=a[i];
		}
		int[] b=new int[n];
		for(int i=0;i<n;i++) b[i]=i+1;
		
		int[] idx={-1,-1};
		for(int i=0;i<n;i++){
			if(a[i]==b[i]) continue;
			if(idx[0]==-1){
				idx[0]=i+1;
			}
			idx[1]=i+1;
		}

		if(idx[0]==-1){
			System.out.println("0 0");
		}else{
			int j=idx[1]-1;
			for(int i=idx[0]-1;i<=idx[1]-1;i++){
				c[j--]=a[i];
			}
			if(Arrays.equals(b,c)){
				System.out.println(idx[0]+" "+idx[1]);
			}else{
				System.out.println("0 0");
			}
		}
	}
	static class FastScanner{
		BufferedReader br;
		StringTokenizer st;
		public FastScanner(){br=new BufferedReader(new InputStreamReader(System.in));}
		String nextToken(){
			while(st==null||!st.hasMoreElements())
				try{st=new StringTokenizer(br.readLine());}catch(Exception e){}
			return st.nextToken();
		}
		int nextInt(){return Integer.parseInt(nextToken());}
		long nextLong(){return Long.parseLong(nextToken());}
		double nextDouble(){return Double.parseDouble(nextToken());}
	}
}
