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;
void fileIO() {}
bool sortbysecdesc(const pair<long long, long long> &a,
                   const pair<long long, long long> &b) {
  return a.second > b.second;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  fileIO();
  long long T;
  T = 1;
  while (T--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (long long &x : a) {
      cin >> x;
    }
    long long start = 0;
    long long l = -1, r = -1;
    for (long long i = 0; i < n; i++) {
      if (a[i] != i + 1) {
        l = i;
        start = a[i];
        break;
      }
    }
    if (start == 0) {
      cout << 0 << " " << 0;
      return 0;
    }
    if (a[start - 1] == l + 1) {
      long long l1 = l;
      long long r = start - 1;
      long long r1 = start - 1;
      while (l1 < r1) {
        long long temp = a[l1];
        a[l1] = a[r1];
        a[r1] = temp;
        l1++;
        r1--;
      }
      for (long long i = 0; i < n; i++) {
        if (a[i] != i + 1) {
          cout << 0 << " " << 0;
          return 0;
        }
      }
      cout << l + 1 << " " << r + 1;
      return 0;
    } else {
      cout << 0 << " " << 0;
    }
  }
  return 0;
}