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;
int a[1001], i, j, n, x, y;
bool b;
int main() {
  b = false;
  cin >> n;
  a[0] = 0;
  a[n + 1] = n + 1;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (i = 1; i <= n; i++) {
    if (a[i] == a[i - 1] + 1)
      continue;
    else {
      x = i;
      for (j = i + 1; j <= n + 1; j++)
        if (a[j] == a[j - 1] - 1)
          continue;
        else if (a[i] == a[j] - 1) {
          y = j - 1;
          break;
          break;
        } else {
          cout << "0 0";
          return 0;
        }
      break;
    }
  }
  for (i = x; i <= (y + x) / 2; i++) swap(a[i], a[y - i + x]);
  for (i = 1; i <= n; i++)
    if (a[i] != a[i - 1] + 1) b = true;
  if (b == false)
    cout << x << " " << y;
  else
    cout << "0 0";
  return 0;
}
