cf 56B

B. Spoilt Permutation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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

 

#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int n,a[1010];
int main()
{
      int l=0,r=0;
      scanf("%d",&n);
      for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
      for(int i=1;i<=n;i++)
            if(a[i]!=i)
            {
                  l=i;
                  break;
            }
      for(int i=n;i>=1;i--)
            if(a[i]!=i)
            {
                  r=i;
                  break;
            }
      for(int i=l;i<r;i++)
      {
            if(a[i]<a[i+1])
            {
                  printf("0 0\n");
                  return 0;
            }
      }
      if(l==0&&r==0)
      {
            printf("0 0\n");
            return 0;
      }
      printf("%d %d",l,r);
      return 0;
}