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
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (a % b == 0) {
    return b;
  } else {
    return gcd(b, a % b);
  }
}
int main() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  for (long long i = 0; i < n; ++i) {
    cin >> v[i];
  }
  long long l = 0, r = 0;
  bool rev = 0;
  for (long long i = 0; i < n; ++i) {
    if (v[i] != i + 1 && !rev) {
      rev = 1;
      l = i + 1;
    } else if (v[i] != i + 1 && rev) {
      r = i + 1;
    }
  }
  vector<long long> p(n);
  for (long long i = 0; i < n; ++i) {
    p[i] = i + 1;
  }
  vector<long long> f(n);
  for (long long i = 0; i < l - 1; ++i) {
    f[i] = i + 1;
  }
  long long g = 1;
  for (long long i = l - 1; i < r; ++i) {
    f[i] = v[r - g];
    g++;
  }
  for (long long i = r; i < n; ++i) {
    f[i] = v[i];
  }
  for (long long i = 0; i < n; ++i) {
    if (f[i] != p[i]) {
      cout << 0 << " " << 0;
      return 0;
    }
  }
  cout << l << " " << r;
}
