#include <bits/stdc++.h>
using namespace std;
int _I() {
  int x;
  scanf("%d", &x);
  return x;
}
long long _L() {
  long long x;
  scanf("%lld", &x);
  return x;
}
long long gcd(long long x, long long y) {
  if (x % y == 0)
    return y;
  else
    return gcd(y, x % y);
}
long long lcm(long long x, long long y) {
  x /= gcd(x, y);
  return x * y;
}
int kdx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int kdy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
bool prime[1000000 + 2];
vector<int> v;
map<int, int> mp;
void sieve() {
  int i, j;
  prime[1] = 1;
  v.push_back(2);
  for (i = 4; i <= 1000000; i += 2) prime[i] = 1;
  for (i = 3; i <= 1000000; i += 2)
    if (!prime[i]) {
      v.push_back(i);
      for (j = i + i + i; j <= 1000000; j += i + i) prime[j] = 1;
    }
}
int main() {
  int i, j, k = 0, n, m, c = _I(), ans = 0;
  int temp = sqrt(c);
  while (1) {
    if (c % temp == 0) {
      ans = c / temp;
      break;
    }
    temp++;
  }
  if (temp > ans) swap(ans, temp);
  printf("%d %d", temp, ans);
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

A big company decided to launch a new series of rectangular displays, and decided that the display must have exactly n pixels. 

Your task is to determine the size of the rectangular display — the number of lines (rows) of pixels a and the number of columns of pixels b, so that:

  * there are exactly n pixels on the display; 
  * the number of rows does not exceed the number of columns, it means a ≤ b; 
  * the difference b - a is as small as possible. 

Input

The first line contains the positive integer n (1 ≤ n ≤ 106) — the number of pixels display should have.

Output

Print two integers — the number of rows and columns on the display. 

Examples

Input

8


Output

2 4


Input

64


Output

8 8


Input

5


Output

1 5


Input

999999


Output

999 1001

Note

In the first example the minimum possible difference equals 2, so on the display should be 2 rows of 4 pixels.

In the second example the minimum possible difference equals 0, so on the display should be 8 rows of 8 pixels.

In the third example the minimum possible difference equals 4, so on the display should be 1 row of 5 pixels.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int _I() {
  int x;
  scanf("%d", &x);
  return x;
}
long long _L() {
  long long x;
  scanf("%lld", &x);
  return x;
}
long long gcd(long long x, long long y) {
  if (x % y == 0)
    return y;
  else
    return gcd(y, x % y);
}
long long lcm(long long x, long long y) {
  x /= gcd(x, y);
  return x * y;
}
int kdx[8] = {-2, -2, -1, -1, 1, 1, 2, 2};
int kdy[8] = {-1, 1, -2, 2, -2, 2, -1, 1};
bool prime[1000000 + 2];
vector<int> v;
map<int, int> mp;
void sieve() {
  int i, j;
  prime[1] = 1;
  v.push_back(2);
  for (i = 4; i <= 1000000; i += 2) prime[i] = 1;
  for (i = 3; i <= 1000000; i += 2)
    if (!prime[i]) {
      v.push_back(i);
      for (j = i + i + i; j <= 1000000; j += i + i) prime[j] = 1;
    }
}
int main() {
  int i, j, k = 0, n, m, c = _I(), ans = 0;
  int temp = sqrt(c);
  while (1) {
    if (c % temp == 0) {
      ans = c / temp;
      break;
    }
    temp++;
  }
  if (temp > ans) swap(ans, temp);
  printf("%d %d", temp, ans);
  return 0;
}

```