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];
int main() {
  int n;
  cin >> n;
  int i, j, x, y, z;
  bool w = false;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++)
    if (a[i] != i) {
      if (a[i] < i) {
        z = i;
        for (j = a[i]; j <= i; j++) {
          if (a[j] != z) break;
          z++;
        }
        if (j == i + 1) {
          for (j = 1; j < a[i]; j++)
            if (a[j] != j) break;
          if (j == a[i]) {
            for (j = i + 1; j <= n; j++)
              if (a[j] != j) break;
            if (j == n + 1) {
              w = true;
              x = a[i];
              y = i;
              break;
            }
          }
        }
      } else {
        z = i;
        for (j = a[i]; j >= i; j--) {
          if (a[j] != z) break;
          z++;
        }
        if (j == i - 1) {
          for (j = 1; j < i; j++)
            if (a[j] != j) break;
          if (j == i) {
            for (j = a[i] + 1; j <= n; j++)
              if (a[j] != j) break;
            if (j == n + 1) {
              w = true;
              x = i;
              y = a[i];
              break;
            }
          }
        }
      }
    }
  if (w)
    cout << x << ' ' << y << endl;
  else
    cout << 0 << ' ' << 0 << endl;
  return 0;
}
