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;
const int INF = INT_MAX;
const long long INFL = LLONG_MAX;
const long double pi = acos(-1);
int dx[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
int a[1000000 + 11];
int Ls[1000000 + 11];
int Rs[1000000 + 11];
int main() {
  ios_base::sync_with_stdio(0);
  cout.precision(15);
  cout << fixed;
  cout.tie(0);
  cin.tie(0);
  int n;
  cin >> n;
  int a[n + 1];
  vector<int> v, w;
  for (int(i) = 1; (i) <= (n); (i)++) {
    cin >> a[i];
  }
  int j = 1;
  int pref = 0;
  while (j <= n) {
    if (a[j] == j)
      pref++;
    else
      break;
    j++;
  }
  j = n;
  int suff = n + 1;
  while (j >= 1) {
    if (a[j] == j)
      suff--;
    else
      break;
    j--;
  }
  int ok = 1;
  for (int k = pref + 1, kk = suff - 1; k <= suff - 1; k++, kk--) {
    if (a[k] == kk)
      ;
    else
      ok = 0;
  }
  if (ok) {
    if (pref + 1 > suff - 1) {
      cout << 0 << " " << 0 << '\n';
      return 0;
    } else
      cout << pref + 1 << " " << suff - 1 << '\n';
    return 0;
  } else {
    cout << 0 << " " << 0 << '\n';
  }
}