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;
int arr[1005], n, et, i, j, l = -1, r = -1, g, c, voy;
bool puede = true;
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) cin >> arr[i];
  if (arr[1] != 1) {
    g = arr[1];
    c = 1;
    l = 1;
    voy = g;
    et = 1;
  }
  for (i = 2; i <= n && puede; i++) {
    switch (et) {
      case 0:
        if (arr[i] != i) {
          l = i;
          c = arr[i - 1] + 1;
          g = arr[i];
          et = 1;
          voy = g;
        }
        break;
      case 1:
        voy--;
        if (arr[i] != voy) {
          if (arr[i] == g + 1) {
            r = i - 1;
            et = 2;
          } else
            puede = false;
        }
        break;
      case 2:
        if (arr[i] != i) puede = false;
        break;
    }
  }
  if (l >= 0 && r == -1 && puede) r = n;
  if (!puede || l == -1)
    cout << "0 0";
  else
    cout << l << " " << r;
  return 0;
}