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 JAVA):

Solution:
import java.io.*;
import java.util.*;
public final class spoilt_perm
{
    static FastScanner sc=new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out=new PrintWriter(System.out);
	
	static boolean check(int[] a,int l,int r,int n)
	{
		int[] b=new int[n+1];
		for(int i=1;i<=n;i++)
		{
			if(i<l || i>r)
			{
				b[i]=a[i];
				if(b[i]!=i)
				{
					return false;
				}
			}
		}
		while(l<r)
		{
			b[l]=a[r];
			b[r]=a[l];
			if(b[l]!=l || b[r]!=r)
			{
				return false;
			}
			l++;
			r--;
		}
		return true;
	}
	
	public static void main(String args[]) throws Exception
	{
		int n=sc.nextInt(),l=0,r=0;
		int[] a=new int[n+1];
		for(int i=1;i<=n;i++)
		{
			a[i]=sc.nextInt();
		}
		outer:
		for(int i=2;i<=n;i++)
		{
			for(int j=1;j<i;j++)
			{
				if(check(a,j,i,n))
				{
					l=j;
					r=i;
					break outer;
				}
			}
		}
		out.println(l+" "+r);
		out.close();
	}
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}