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 long long int M = 1000000007;
const long long int N = 1e5 + 1;
double PI = 2 * acos(0.0);
vector<long long int> isprime(N, true);
vector<long long int> prime;
vector<long long int> spf(N);
void seive() {
  isprime[0] = isprime[1] = false;
  for (long long int i = 2; i < N; i++) {
    if (isprime[i]) {
      prime.push_back(i);
      spf[i] = i;
    }
    for (long long int j = 0; j < (long long int)prime.size() &&
                              i * prime[j] < N && prime[j] <= spf[i];
         j++) {
      isprime[i * prime[j]] = false;
      spf[i * prime[j]] = prime[j];
    }
  }
}
vector<long long int> getPrimeFacts(long long int x) {
  vector<long long int> ret;
  while (x != 1) {
    ret.push_back(spf[x]);
    x = x / spf[x];
  }
  return ret;
}
long long int fast_pow(long long int a, long long int p) {
  long long int res = 1;
  while (p) {
    if (p % 2 == 0) {
      a = a * 1ll * a % M;
      p /= 2;
    } else {
      res = res * 1ll * a % M;
      p--;
    }
  }
  return res;
}
long long int fact(long long int n) {
  long long int res = 1;
  for (long long int i = 1; i <= n; i++) {
    res = res * 1ll * i % M;
  }
  return res;
}
long long int ncr(long long int n, long long int k) {
  return fact(n) * 1ll * fast_pow(fact(k), M - 2) % M * 1ll *
         fast_pow(fact(n - k), M - 2) % M;
}
bool isPowerOfTwo(long long int n) {
  if (n == 0) return false;
  return (ceil(log2(n)) == floor(log2(n)));
}
long long int strtoint(string s) {
  stringstream geek(s);
  long long int x = 0;
  geek >> x;
  return x;
}
void getVec(vector<long long int>& v, long long int n) {
  for (long long int i = 0; i < n; i++) {
    long long int a;
    cin >> a;
    v.push_back(a);
  }
}
long long int ask(long long int l, long long int r) {
  if (l >= r) return -1;
  cout << "? " << l + 1 << ' ' << r + 1 << endl;
  long long int ans;
  cin >> ans;
  return ans - 1;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a;
  getVec(a, n);
  long long int l = -1, r = -1;
  for (long long int i = 0; i < n; i++) {
    if (a[i] != i + 1) {
      l = i;
      break;
    }
  }
  if (l == -1) {
    cout << "0 0";
    return;
  }
  for (long long int i = n - 1; i >= 0; i--) {
    if (a[i] != i + 1) {
      r = i;
      break;
    }
  }
  long long int t = l;
  for (long long int i = r; i >= l; i--) {
    if (a[i] != t + 1) {
      cout << "0 0";
      return;
    }
    t++;
  }
  cout << l + 1 << " " << r + 1;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) solve();
}