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 mx, i, j, l, r, p2, p, n, a[10110];
int main() {
  cin >> n;
  for (i = 1; i <= n; i++) cin >> a[i];
  i = 2;
  p = a[1];
  mx = 0;
  while (i <= n) {
    j = i + 1;
    if (p - 1 == a[i]) {
      j = i + 1;
      p2 = a[i];
      while (j <= n && p2 - 1 == a[j]) {
        p2 = a[j];
        j++;
      }
      if (j - i > mx) {
        mx = j - i;
        l = i - 1;
        r = j - 1;
      }
    }
    p = a[i];
    i = j;
  }
  p = a[r];
  for (i = l; i <= r; i++) {
    a[i] = p;
    p++;
  }
  bool boo = true;
  for (i = 2; i <= n; i++)
    if (a[i] != a[i - 1] + 1) boo = false;
  if (!boo) {
    cout << "0 0";
    return 0;
  }
  printf("%d %d", l, r);
  return 0;
}