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 a[100001];
int main() {
  int n, i, j, r, s, cnt, w, f, g, t;
  scanf("%d", &n);
  int ara[100000];
  for (i = 0; i < n; i++) {
    scanf("%d", &ara[i]);
  }
  f = 0;
  for (i = 0; i < n;) {
    if (ara[i] != i + 1) {
      f++;
      s = ara[i];
      g = i;
      t = i + 1;
      if (ara[s - 1] != i + 1) {
        printf("0 0");
        return 0;
      } else {
        w = t;
        for (j = i;; j++) {
          if ((ara[g] - t) % 2 == 0 && ara[ara[j]] == j) {
            break;
          } else if ((ara[g] - t) % 2 == 1 && ara[j] - w == 1) {
            break;
          }
          if (ara[ara[j] - 1] != w) {
            printf("0 0");
            return 0;
          }
          w = w + 1;
        }
      }
      i = ara[i];
    } else {
      i = i + 1;
    }
  }
  if (f == 0 || f > 1) {
    printf("0 0");
    return 0;
  } else {
    printf("%d %d", t, ara[g]);
  }
  return 0;
}