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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long INF = 1e18L;
const double eps = 1e-9;
const long long mod = 1e9 + 7;
const double pi = 2.0 * acos(0.0);
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  bool ok = true;
  for (long long i = 1; i <= n - 1; i++) {
    if (a[i] < a[i - 1]) ok = false;
  }
  if (ok) {
    cout << "0 0\n";
    return 0;
  }
  long long x = -1, y = -1;
  for (long long i = 0; i < n - 1; i++) {
    if (a[i] != i + 1) {
      x = i;
      y = i;
      while (y < n - 1 && a[y] == a[y + 1] + 1) {
        y++;
      }
      break;
    }
  }
  ok = true;
  for (long long i = 0; i < x - 1; i++) {
    if (a[i] != a[i + 1] - 1) {
      ok = false;
    }
  }
  for (long long i = x; i < y; i++) {
    if (a[i] != a[i + 1] + 1) {
      ok = false;
    }
  }
  for (long long i = y + 1; i < n - 1; i++) {
    if (a[i] != a[i + 1] - 1) {
      ok = false;
    }
  }
  if (ok && x < y) {
    cout << x + 1 << " " << y + 1 << "\n";
  } else {
    cout << "0 0\n";
  }
  return 0;
}