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;
priority_queue<int, vector<int>, greater<int> > Q;
const int INF = 1e9 + 9;
long int pow(long int a, long int b, int m) {
  a %= m;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
int main() {
  int n;
  cin >> n;
  int ar[n + 5];
  for (int i = 1; i <= n; i++) {
    cin >> ar[i];
  }
  int ans[n + 5];
  for (int i = 1; i <= n; i++) {
    ans[i] = i;
  }
  int k = 0, c = 0;
  int ind = 0;
  int si = 0;
  for (int i = 1; i <= n; i++) {
    if (ar[i] != ans[i]) {
      c++;
      si = i;
      for (k = i; k <= n; k++, i++) {
        if (ar[k + 1] > ar[k]) {
          ind = k;
          break;
        }
      }
    }
  }
  if (c > 1) {
    cout << 0 << " " << 0 << endl;
  }
  bool flag = true;
  for (int j = k + 1; j <= ind; j++) {
    if (ar[j] != ar[j - 1] - 1) {
      flag = false;
      cout << 0 << " " << 0 << endl;
      return 0;
    }
  }
  if (flag and c == 1) {
    cout << si << " " << ind << endl;
  }
  if (c == 0) {
    cout << 0 << " " << 0 << endl;
  }
}
