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 res, a, b;
void solve(vector<long long> v, long long i, long long x, long long cur) {
  long long j, k, n = v.size();
  if (i == n) {
    long long x1 = max(cur, x / cur);
    if (res > x1) {
      res = x1;
      a = cur;
      b = x / cur;
    }
    return;
  }
  solve(v, i + 1, x, cur);
  solve(v, i + 1, x, cur * v[i]);
}
int main() {
  long long i, j, k, n;
  cin >> n;
  map<long long, long long> mp;
  long long x = n;
  while (x % 2 == 0) {
    mp[2]++;
    x /= 2;
  }
  for (i = 3; i * i <= x; i += 2) {
    while (x % i == 0) {
      mp[i]++;
      x /= i;
    }
  }
  if (x > 1) mp[x]++;
  vector<long long> v;
  for (auto it = mp.begin(); it != mp.end(); it++) {
    v.push_back(it->first);
  }
  for (i = 0; i < v.size(); i++) {
    v[i] = pow(v[i], mp[v[i]]);
  }
  res = n;
  a = n;
  b = 1;
  solve(v, 0LL, n, 1LL);
  cout << a << ' ' << b;
}