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;
void read_file(bool outToFile = true) {}
int main() {
  read_file();
  int n;
  while (cin >> n) {
    vector<int> A(n);
    for (int i = 0; i < n; i++) cin >> A[i];
    int x, y;
    x = y = -1;
    for (int i = 0; i < n - 1 && x == -1; i++)
      if (A[i] > A[i + 1]) x = i;
    if (x == -1) {
      cout << "0 0\n";
      continue;
    }
    for (int i = x + 1; i < n - 1 && y == -1; i++)
      if (A[i] < A[i + 1]) y = i;
    if (y == -1) y = n - 1;
    reverse(A.begin() + x, A.begin() + y + 1);
    bool ok = true;
    for (int i = 0; i < n - 1; i++)
      if (A[i] > A[i + 1]) ok = false;
    if (!ok) {
      cout << "0 0\n";
      continue;
    }
    cout << x + 1 << ' ' << y + 1 << '\n';
  }
}
