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 n;
vector<int> v;
bool isO() {
  for (int i = 0; i < n; ++i)
    if (v[i] != i + 1) return false;
  return true;
}
void solve() {
  int a, b;
  a = b = -1;
  for (int i = 0; i < n; ++i) {
    if (i + 1 != v[i]) {
      a = i;
      break;
    }
  }
  for (int j = n - 1; j >= 0; --j) {
    if (j + 1 != v[j]) {
      b = j;
      break;
    }
  }
  int inicio = b + 1;
  bool ret = true;
  for (int i = a; i <= b; ++i) {
    if (inicio != v[i]) {
      ret = false;
      break;
    }
    --inicio;
  }
  if (!ret)
    cout << "0 0" << endl;
  else
    cout << (a + 1) << " " << (b + 1) << endl;
}
int main() {
  int tmp;
  while (cin >> n && n > 0) {
    v.clear();
    for (int i = 0; i < n; ++i) {
      cin >> tmp;
      v.push_back(tmp);
    }
    if (isO()) {
      cout << "0 0" << endl;
      continue;
    }
    solve();
  }
  return 0;
}