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>
#pragma comment(linker, "/STACK:136777216")
using namespace std;
int bits(int x) { return x == 0 ? 0 : 1 + bits(x & (x - 1)); }
const double PI = acos(-1.0);
const double eps = 1e-9;
const int INF = 1000000000;
string nextString() {
  char buf[10000000];
  gets(buf);
  return buf;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  scanf("%d\n", &n);
  vector<int> in(n);
  for (int(i) = 0; (i) < (n); (i)++) scanf("%d", &in[i]);
  int l = -1;
  for (int(i) = 0; (i) < (n); (i)++)
    if (in[i] != i + 1) {
      l = i;
      break;
    }
  if (l == -1) {
    puts("0 0");
    return 0;
  }
  int r = in[l];
  for (int i = l, j = r; i < r; i++, j--)
    if (in[i] != j) {
      puts("0 0");
      return 0;
    }
  for (int(i) = (r); (i) < (n); (i)++)
    if (in[i] != i + 1) {
      puts("0 0");
      return 0;
    }
  printf("%d %d\n", l + 1, r);
  return 0;
}