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;
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-ffloat-store")
#pragma GCC optimize("-fno-defer-pop")
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long int t = power(a, b / 2, m) % m;
  t = (t * t) % m;
  if (b & 1) t = ((t % m) * (a % m)) % m;
  return t;
}
long long int modInverse(long long int a, long long int m) {
  return power(a, m - 2, m);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, l, n;
  cin >> n;
  vector<long long int> a, b;
  for (i = 1; i <= n; i++) {
    cin >> k;
    a.push_back(k);
  }
  b = a;
  sort(b.begin(), b.end());
  if (a == b) {
    cout << "0 0\n";
    return 0;
  }
  long long int mni = 1e18, mxi = 0;
  for (i = 0; i < n; i++) {
    if (a[i] != b[i]) {
      mxi = max(mxi, i);
      mni = min(mni, i);
    }
  }
  reverse(a.begin() + mni, a.begin() + mxi + 1);
  if (a == b) {
    cout << mni + 1 << " " << mxi + 1 << "\n";
  } else {
    cout << "0 0";
  }
  return 0;
}
