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 prime[1000000 + 1];
vector<long> v;
void sieve() {
  memset(prime, true, sizeof(prime));
  for (long p = 2; p * p <= 1000000; p++) {
    if (prime[p] == true) {
      for (long i = p * p; i <= 1000000; i += p) prime[i] = false;
    }
  }
  for (int i = 0; i <= 1000000; i++)
    if (prime[i]) v.push_back(i);
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long power(int v, int a) {
  long long prod = 1;
  for (int i = 0; i < a; i++) prod *= v;
  return prod;
}
long double dist(long long a, long long b, long long c, long long d) {
  double t = (a - c) * (a - c);
  double p = (b - d) * (b - d);
  return sqrt(t + p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T = 1;
  while (T--) {
    int n, l, r, cnt = 0;
    cin >> n;
    int a[n + 1];
    a[0] = 0;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) {
      if (a[i] != i && cnt != 0) {
        cout << "0 0";
        return 0;
      } else if (a[i] != i) {
        cnt++;
        l = i;
        while (i <= n && a[i] != l && a[i] > a[i + 1]) i++;
        if (a[i] != l || a[l] != i) {
          cout << "0 0";
          return 0;
        }
        r = i;
      }
    }
    if (cnt != 1)
      cout << "0 0";
    else
      cout << l << " " << r;
  }
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
}