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 main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, l = 0, r = 0, cnt = 0, c = 0, j = 0;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n - 1; ++i) {
    if (a[i + 1] < a[i] && a[i] - a[i + 1] == 1) l = i + 1;
    j = i + 1;
    while (a[j] < a[i] && i < n && j < n) {
      cnt++;
      j++;
    }
    if (cnt != 0) break;
  }
  reverse(a + l - 1, a + l + cnt);
  if (is_sorted(a, a + n))
    cout << l << " " << l + cnt;
  else
    cout << 0 << " " << 0;
  cerr << 1.0 * clock() / CLOCKS_PER_SEC << "s";
  return 0;
}