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>
using namespace std;
template <class T>
void show(T a, int n) {
  for (int i = 0; i < n; ++i) cout << a[i] << ' ';
  cout << endl;
}
template <class T>
void show(T a, int r, int l) {
  for (int i = 0; i < r; ++i) show(a[i], l);
  cout << endl;
}
const int N = 128 * 2;
const int M = 5000;
const int oo = 10000 * 10000 * 10;
int a[2000];
int n;
bool ok() {
  for (int i = 1; i < n; ++i)
    if (a[i] < a[i - 1]) return false;
  return true;
}
void solve() {
  int i, j, k;
  if (ok()) {
    puts("0 0");
    return;
  }
  i = 0, j = n - 1;
  while (a[i] == i + 1) ++i;
  while (a[j] == j + 1) --j;
  int lf = i;
  int ri = j;
  while (i < j) {
    swap(a[i], a[j]);
    ++i;
    --j;
  }
  if (ok()) {
    printf("%d %d\n", lf + 1, ri + 1);
    return;
  }
  puts("0 0");
  return;
}
int main() {
  int i, j, cas = 0;
  cin >> n;
  for (i = 0; i < n; ++i) cin >> a[i];
  solve();
  return 0;
}