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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long int power(long long int x, long long int y) {
  long long int res = 1;
  while (y) {
    if (y % 2 == 0) {
      x = (x * x);
      y /= 2;
    } else {
      res = (res * x);
      y--;
    }
  }
  return res;
}
void sieve() {
  vector<bool> prime(1e6 + 1, true);
  prime[0] = prime[1] = false;
  for (long long int i = 2; i * i <= 1e6; i++) {
    if (prime[i]) {
      for (long long int j = i * i; j <= 1e6; j += i) prime[j] = false;
    }
  }
}
bool is_Balanced_Parenthesis(string s) {
  stack<char> st;
  for (auto it : s) {
    if (it == '(' || it == '{' || it == '[') {
      st.push(it);
      continue;
    }
    if (st.empty()) return false;
    if (it == ')') {
      char ch = st.top();
      st.pop();
      if (ch != '(') return false;
    }
    if (it == ']') {
      char ch = st.top();
      st.pop();
      if (ch != '[') return false;
    }
    if (it == '}') {
      char ch = st.top();
      st.pop();
      if (ch != '{') return false;
    }
  }
  return st.empty();
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a(n + 1);
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  long long int x = 0, y = 0, min = 0;
  for (long long int i = 1; i <= n; i++) {
    if (a[i] != i) {
      min = i;
      break;
    }
  }
  if (!min) {
    cout << x << " " << y << "\n";
    return;
  }
  x = min;
  y = a[min];
  reverse(a.begin() + min, a.begin() + 1 + y);
  for (long long int i = 1; i <= n; i++) {
    if (i != a[i]) {
      cout << "0 0"
           << "\n";
      return;
    }
  }
  cout << x << " " << y << "\n";
  return;
}
void c_p_c() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int32_t main() {
  c_p_c();
  solve();
  return 0;
}