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>
#pragma GCC optimize("O2")
using namespace std;
int main() {
  cin.tie(NULL);
  cout.tie(NULL);
  {
    long long int n;
    cin >> n;
    long long int a = 1, b = 1;
    while (n % 2 == 0) {
      a *= 2;
      n /= 2;
    }
    vector<long long int> v;
    for (long long int i = 3; i * i <= n; i += 2) {
      long long int cur = 1;
      while (n % i == 0) {
        cur *= i;
        n /= i;
      }
      b *= cur;
      if (cur > 1) v.push_back(cur);
    }
    if (n > 1) {
      v.push_back(n);
      b *= n;
    }
    long long int x = 0;
    for (int i = 0; i < v.size(); i++) x = (x << 1) | 1;
    long long int ans = LONG_LONG_MAX, fa = a, fb = b;
    for (int i = 0; i < x; i++) {
      long long int temp = 1;
      for (int j = 0; j < v.size(); j++) {
        if ((i & (1 << j)) != 0) temp *= v[j];
      }
      if (ans > fabs(temp * a - (b / temp))) {
        ans = fabs(temp * a - (b / temp));
        fa = temp * a;
        fb = b / temp;
      }
    }
    cout << fa << " " << fb << endl;
  }
}