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 mod1 = static_cast<long long>(1073741824);
const long long mod2 = static_cast<long long>(1000000007);
const long long INF = 1ll * 1000 * 1000 * 1000 * 1000 * 1000 * 1000 + 7;
const long double ldOne = 1.0;
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long exp(long long x, long long n) {
  if (n == 0)
    return 1;
  else if (n & 1)
    return x * exp(x * x, (n - 1) / 2);
  else
    return exp(x * x, n / 2);
}
bool isPrime(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
vector<long long> SieveOfEratosthenes(long long n) {
  vector<bool> prime(n + 1, true);
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p]) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  vector<long long> v;
  for (long long i = 2; i <= n; i++) {
    if (prime[i]) v.push_back(i);
  }
  return v;
}
map<long long, long long> PrimeFactorization(long long n) {
  vector<long long> v = SieveOfEratosthenes(static_cast<long long>(sqrt(n)));
  map<long long, long long> m;
  for (unsigned long long i = 0; i < v.size(); i++) {
    if (n % v[i] == 0) {
      n /= v[i];
      m[v[i]]++;
      i--;
    }
  }
  return m;
}
long long nCr(long long n, long long r) {
  vector<long long> c(r + 1, 0);
  c[0] = 1;
  for (long long i = 0; i < n + 1; i++) {
    for (long long j = min(i, r); j > 0; j--) c[j] = c[j - 1] + c[j];
  }
  return c[r];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  long long ax[1555];
  cin >> n;
  for (long long i = 1; i <= n; ++i) cin >> ax[i];
  long long i = 1;
  long long j = n;
  for (; i <= n; ++i)
    if (ax[i] != i) break;
  for (; j >= 1; --j)
    if (ax[j] != j) break;
  if (i == n + 1)
    cout << 0 << " " << 0 << "\n";
  else {
    for (long long a = i, b = j; a <= b; ++a, --b)
      if (ax[a] != b || ax[b] != a) {
        cout << 0 << " " << 0 << "\n";
        exit(0);
      }
    cout << i << " " << j << "\n";
  }
  cerr << "Time : " << 1000 * (long double)clock() / (long double)CLOCKS_PER_SEC
       << "ms\n";
  ;
  return 0;
}