Today, Osama gave Fadi an integer X, and Fadi was wondering about the minimum possible value of max(a, b) such that LCM(a, b) equals X. Both a and b should be positive integers.

LCM(a, b) is the smallest positive integer that is divisible by both a and b. For example, LCM(6, 8) = 24, LCM(4, 12) = 12, LCM(2, 3) = 6.

Of course, Fadi immediately knew the answer. Can you be just like Fadi and find any such pair?

Input

The first and only line contains an integer X (1 ≤ X ≤ 10^{12}).

Output

Print two positive integers, a and b, such that the value of max(a, b) is minimum possible and LCM(a, b) equals X. If there are several possible such pairs, you can print any.

Examples

Input


2


Output


1 2


Input


6


Output


2 3


Input


4


Output


1 4


Input


1


Output


1 1
#include <bits/stdc++.h>
using namespace std;
template <class T>
int check_mod(T a, T b) {
  return (a % b) ? 1 : 0;
}
template <class T>
void maxi(T &a, T b) {
  a = max(a, b);
  return;
}
template <class T>
void mini(T &a, T b) {
  a = min(a, b);
  return;
}
bool is_prime(int a) {
  if (a == 2) return true;
  if (!(a & 1) || a == 1) return false;
  for (int i = 3; i * i <= a; i += 2)
    if (!(a % i)) return false;
  return true;
}
bool is_prime(long long a) {
  if (a == 2) return true;
  if (!(a & 1) || a == 1) return false;
  for (long long i = 3; i * i <= a; i += 2)
    if (!(a % i)) return false;
  return true;
}
bool is_palindrome(string s) {
  int len = (int)s.length();
  int mid = len >> 1;
  for (int i = 0; i < mid; i++)
    if (s[i] != s[len - i - 1]) return false;
  return true;
}
int gcd(int a, int b) {
  while (b) {
    int mod = a % b;
    a = b;
    b = mod;
  }
  return a;
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
int binpow(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
int binpow(int a, int b, int MODULO) {
  int res = 1;
  while (b) {
    if (b & 1) {
      res *= a;
      res %= MODULO;
    }
    a *= a;
    a %= MODULO;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) {
  while (b) {
    long long mod = a % b;
    a = b;
    b = mod;
  }
  return a;
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res *= a;
    a *= a;
    b >>= 1;
  }
  return res;
}
long long binpow(long long a, long long b, long long MODULO) {
  long long res = 1;
  while (b) {
    if (b & 1) {
      res *= a;
      res %= MODULO;
    }
    a *= a;
    a %= MODULO;
    b >>= 1;
  }
  return res;
}
string binary(int N) {
  if (!N) return "0";
  string bin = "";
  while (N) {
    bin += '0' + (N & 1);
    N >>= 1;
  }
  reverse((bin).begin(), (bin).end());
  return bin;
}
string binary(long long N) {
  if (!N) return "0";
  string bin = "";
  while (N) {
    bin += '0' + (N & 1);
    N >>= 1;
  }
  reverse((bin).begin(), (bin).end());
  return bin;
}
const int INF = 2147483547;
const int MODULO = 1000000007;
const int _MODULO = 998244353;
const long long INF64 = 9223372036854775707;
const long long MODULO64 = 1000000007;
const long long _MODULO64 = 998244353;
const double epsilon = 10e-6;
const double PI = acos(-1);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long x;
  cin >> x;
  pair<long long, long long> ans = {INF64, INF64};
  for (long long a = 1; a * a <= x; ++a) {
    long long b = x / a;
    if (a * b != x) continue;
    if (lcm(a, b) != x) continue;
    if (b < ans.second) ans = make_pair(a, b);
  }
  cout << ans.first << ' ' << ans.second << '\n';
  return 0;
}
