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>
const double pi = acos(-1.0);
using namespace std;
inline long long MAX2(long long a, long long b) {
  return (a) > (b) ? (a) : (b);
}
inline long long MAX3(long long a, long long b, long long c) {
  return (a) > (b) ? ((a) > (c) ? (a) : (c)) : ((b) > (c) ? (b) : (c));
}
inline long long MIN2(long long a, long long b) {
  return (a) < (b) ? (a) : (b);
}
inline long long MIN3(long long a, long long b, long long c) {
  return (a) < (b) ? ((a) < (c) ? (a) : (c)) : ((b) < (c) ? (b) : (c));
}
int mod = 1e9 + 7;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long powmod(long long a, long long b) {
  long long res = 1;
  if (a >= mod) a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    if (res >= mod) res %= mod;
    a = a * a;
    if (a >= mod) a %= mod;
  }
  return res;
}
unsigned long int setbits(long long n) {
  unsigned long int count = 0;
  while (n) {
    n &= (n - 1);
    count++;
  }
  return count;
}
int modInverse(int A, int M) { return powmod(A, M - 2); }
bool is_prime(long long n) {
  if (n == 1) {
    return false;
  }
  long long i = 2;
  while (i * i <= n) {
    if (n % i == 0) {
      return false;
    }
    i += 1;
  }
  return true;
}
long long lcm(long long a, long long b) { return (a * b) / gcd(a, b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  if (is_prime(n)) {
    cout << 1 << " " << n << '\n';
    return 0;
  }
  int end = sqrt(n);
  long long second = LLONG_MAX, en = LLONG_MAX;
  for (int i = 1; i <= end; i++) {
    if (n % i == 0) {
      long long res1 = i;
      long long res2 = n / i;
      if (lcm(res1, res2) == n) {
        second = res1;
        en = res2;
      }
    }
  }
  cout << second << " " << en << '\n';
  return 0;
}
