Spoilt Permutation

http://codeforces.com/problemset/problem/56/B

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

Code:

 #include<iostream>
using namespace std;
int main()
{
    int x,y,n,a[1001];
    cin>>n;
    for(int i=0;i<n;i++)
    {
 cin>>a[i];
    }
    x=0;
    y=n-1;
    while(a[x]==x+1)
    {
 x++;
    }
    while(a[y]==y+1)
    {
 y--;
    }
    if(x>y)
    {
 cout<<"0 0\n";
    }
    else
    {
 for(int i=x;i<=y;i++)
 {
     if(a[i]!=y+x+1-i)
     {
  cout<<"0 0\n";
  return 0;
     }
 }
 cout<<x+1<<" "<<y+1<<endl;
    }
    return 0;
}