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.*;
import java.util.*;

public class Solution {

    StreamTokenizer in;
    PrintWriter out;

    public static void main(String[] args) throws Exception {
        new Solution().run();
    }

    public void run() throws Exception {
        in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        out = new PrintWriter(new OutputStreamWriter(System.out));
        solve();
        out.flush();
    }

    int nextInt() throws Exception {
        in.nextToken();
        return (int) in.nval;
    }

    String next() throws Exception {
        in.nextToken();
        return in.sval;
    }

    public void solve() throws Exception {
    int n =nextInt();
    int beg=1;
    int tail=n;
    int secbeg=0;
    int sectail=0;
    int [] per = new int [n+2];
    for (int i=0; i<n;i++)
    {
     per[i]=nextInt();
    }
    if (n==1) {out.println("0 0");out.flush();return;}
    boolean rev=true;
    for (int i=0;i<n-1;i++)
    {
        if (per[i]<per[i+1]) rev=false;
    }
    if (rev) {out.println("1 "+n);out.flush();return;}
    boolean flag=true;
    for (int i=0; i<n-1;i++)
    {
        if (flag)
        {
        if (per[i+1]-per[i]!=1) {
            sectail=per[i+1];
            int j=sectail-1;
            if (per[j]-per[i]!=1){ out.println("0 0"); out.flush();return;}
            for (j=sectail-1; j>i+1; j--)
               if (per[j-1]-per[j]!=1){
                   out.println("0 0");
                   out.flush();
                   return;
               }
            secbeg=per[sectail-1];
            flag=false;
            i=sectail;
        }} else if (per[i+1]-per[i]!=1)
        {
            secbeg=0;
            sectail=0;
            break;
        }
    }
     out.println(secbeg+" "+sectail);
    }
}
