Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long a, long long b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long expo(long long a, long long b, long long mod) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
vector<int> sieve(int n) {
  int *arr = new int[n + 1]();
  vector<int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
void extendgcd(long long a, long long b, long long *v) {
  if (b == 0) {
    v[0] = 1;
    v[1] = 0;
    v[2] = a;
    return;
  }
  extendgcd(b, a % b, v);
  long long x = v[1];
  v[1] = v[0] - v[1] * (a / b);
  v[0] = x;
  return;
}
long long mminv(long long a, long long b) {
  long long arr[3];
  extendgcd(a, b, arr);
  return arr[0];
}
long long mminvprime(long long a, long long b) { return expo(a, b - 2, b); }
bool revsort(long long a, long long b) { return a > b; }
void swap(int &x, int &y) {
  int temp = x;
  x = y;
  y = temp;
}
long long combination(long long n, long long r, long long m, long long *fact) {
  long long val1 = fact[n];
  long long val2 = mminvprime(fact[r], m);
  long long val3 = mminvprime(fact[n - r], m);
  return ((val1 * val2) % m * val3) % m;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long x;
  cin >> x;
  long long val = x;
  for (long long i = 2; i <= sqrt(x); i++)
    if (x % i == 0)
      if (x / gcd(i, x / i) == x) val = min(val, max(i, x / i));
  cout << x / val << " " << val << endl;
  return 0;
}