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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
bool symb(char b) {
  switch (b) {
    case '1':
      return true;
      break;
    case '2':
      return true;
      break;
    case '3':
      return true;
      break;
    case '4':
      return true;
      break;
    case '5':
      return true;
      break;
    case '6':
      return true;
      break;
    case '7':
      return true;
      break;
    case '8':
      return true;
      break;
    case '9':
      return true;
      break;
    case '0':
      return true;
      break;
    default:
      return false;
      break;
  }
}
int change(string s) {
  int x = 1, c = 0;
  for (int j = s.size() - 1; j >= 0; j--) {
    switch (s[j]) {
      case '1':
        c += x;
        break;
      case '2':
        c += x * 2;
        break;
      case '3':
        c += x * 3;
        break;
      case '4':
        c += x * 4;
        break;
      case '5':
        c += x * 5;
        break;
      case '6':
        c += x * 6;
        break;
      case '7':
        c += x * 7;
        break;
      case '8':
        c += x * 8;
        break;
      case '9':
        c += x * 9;
        break;
      case '0':
        c += x * 0;
        break;
    }
    x *= 10;
  }
  return c;
}
int main() {
  int a[1002], n, c = 0, k = 0, b = 0, e = 0;
  bool q = true, sl = true, sl1 = true, pr = true, norm = true;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == i && q)
      c++;
    else {
      if (q && b == 0) {
        b = i;
        k = a[i] + i;
        q = false;
      } else {
        if (k == a[i] + i) {
          e = i;
          ;
        } else {
          if (a[i] == i)
            continue;
          else {
            norm = false;
            break;
          }
        }
      }
    }
  }
  if (!norm)
    cout << 0 << " " << 0;
  else if (c == n)
    cout << 0 << " " << 0;
  else
    cout << b << " " << e;
  return 0;
}