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;
const long long N = 2e5 + 1298;
long long solve() {
  long long n;
  cin >> n;
  long long arr[n];
  for (long long i = 0; i < n; ++i) cin >> arr[i];
  long long l = -1, r = -1;
  for (long long i = 0; i < n; i++) {
    if (i == 0 && arr[i] != 1) {
      l = i;
      break;
    } else if (i && arr[i - 1] > arr[i]) {
      l = i - 1;
      break;
    }
  }
  for (long long i = n - 1; i >= 0; i--) {
    if (i == n - 1 && arr[i] != n) {
      r = i;
      break;
    } else if (i < n - 1 && arr[i] > arr[i + 1]) {
      r = i + 1;
      break;
    }
  }
  long long L = l, R = r;
  long long req[n];
  for (long long i = 0; i < n; ++i) req[i] = i + 1;
  while (r > l) swap(arr[r], arr[l]), r--, l++;
  l = L, r = R;
  for (long long i = 0; i < n; ++i)
    if (req[i] != arr[i]) return cout << 0 << " " << 0, 0;
  return cout << l + 1 << " " << r + 1, 0;
  return 0;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve(), cout << '\n';
  cerr << (float)clock() / CLOCKS_PER_SEC << " secs\n";
  return 0;
}