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;
bool sortbysecdesc(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second > b.second;
}
long long i, j;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(NULL), cout.tie(NULL);
  long long t = 1;
  while (t--) {
    long long n, count = 0, flag = 0, loc;
    cin >> n;
    vector<long long> v(n + 1);
    v[0] = 0;
    for (i = 1; i <= n; i++) {
      cin >> v[i];
      if (v[i] == i) {
        count++;
      }
    }
    if (count == n) {
      cout << "0 0";
      return 0;
    } else {
      for (i = 1; i <= n; i++) {
        if (v[i] != i) {
          loc = v[i];
          break;
        }
      }
      long long loc2 = i;
      reverse(v.begin() + loc2, v.begin() + loc + 1);
      for (i = 1; i <= n; i++) {
        if (v[i] != i) {
          cout << "0 0";
          return 0;
        }
      }
      cout << loc2 << " " << loc;
    }
  }
}