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>
int main() {
  long n;
  while (scanf("%ld", &n) != EOF) {
    long i, a[100000] = {0}, k = 0, l = 0, b[100000] = {0}, s = 0, p = 0,
            a1[100000] = {0}, z, y;
    for (i = 0; i < n; i++) scanf("%ld", &a[i]);
    for (i = 1; i <= n; i++) {
      a1[k++] = i;
    }
    for (i = 0; i < n; i++) {
      if (a[i] != a1[i]) {
        z = i;
        break;
      }
    }
    for (i = n - 1; i > 0; i--) {
      if (a[i] != a1[i]) {
        y = i;
        break;
      }
    }
    for (i = z; i < y; i++) {
      if (a[i] - a[i + 1] == 1) {
        p = 1;
      } else {
        p = 0;
        break;
      }
    }
    if (p == 1)
      printf("%ld %ld\n", a[y], a[z]);
    else if (p == 0)
      printf("0 0\n");
  }
  return 0;
}