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 double PI = 2 * acos(0.0);
const int Mod = 1E9 + 7;
inline void yes() { std::cout << "Yes\n"; }
inline void no() { std::cout << "No\n"; }
using namespace std;
long long int isPrime(long long int num) {
  long long int flag = 0;
  for (long long int i = 1; i <= num; i++) {
    if (num % i == 0) {
      flag++;
      if (flag > 3) break;
    }
  }
  if (flag == 3)
    return 1;
  else
    return 0;
}
long long int binomialCoeff(long long int n, long long int k) {
  long long int C[k + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = min(i, k); j > 0; j--) C[j] = C[j] + C[j - 1];
  }
  return C[k];
}
long long int factorial(long long int n) {
  if (n == 0) return 1;
  return n * factorial(n - 1);
}
bool isPalindrome(string str) {
  int l = 0;
  int h = str.length() - 1;
  while (h > l) {
    if (str[l++] != str[h--]) {
      return false;
    }
  }
  return true;
}
bool check(long long int n) {
  double y = sqrt(n);
  if (y - floor(y) == 0)
    return 1;
  else
    return 0;
}
bool isprime(long long int N) {
  if (N < 2 || (!(N & 1) && N != 2)) return 0;
  for (int i = 3; i * i <= N; i += 2) {
    if (!(N % i)) return 0;
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int x, y, cnt = 0, l, o, t, n, sum = 0, lasum = 0, ans = 0, las = 0,
                      flag = 0, ma = LLONG_MAX, av1, av2, sum1 = 0, sum2 = 0;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = (0); i < (n); i++) {
    cin >> v[i];
  }
  long long int start = 0, end = 0, now;
  if (is_sorted(v.begin(), v.end())) {
    cout << "0 0";
    return 0;
  }
  vector<long long int> temp;
  temp = v;
  reverse(temp.begin(), temp.end());
  if (is_sorted(temp.begin(), temp.end())) {
    cout << "1 " << v.size();
    return 0;
  }
  for (long long int i = (0); i < (n); i++) {
    if (v[i] != i + 1 && start == 0) {
      start = i;
      now = i + 1;
    }
    if (start != 0 && v[i] == now) {
      end = i + 1;
      break;
    }
  }
  reverse(v.begin() + start, v.begin() + end);
  if (is_sorted(v.begin(), v.end()))
    cout << now << " " << end;
  else
    cout << "0 0";
}