#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long ison(long long mask, long long pos) { return (mask & (1 << pos)); }
long long cbit(long long n) {
  long long k = 0;
  while (n) n &= (n - 1), k++;
  return k;
}
long long nbit(long long n) {
  long long k = 0;
  while (n) n /= 2, k++;
  return k;
}
long long mod = 1e9 + 7;
int sgn(long long x) { return x < 0 ? -1 : !!x; }
const double eps = 1e-7;
const double pi = acos(-1.0);
inline void inc(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline void dec(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
inline void chmax(long long &x, long long y) {
  if (y > x) x = y;
}
inline void mulm(long long &x, long long y) {
  x *= y;
  if (x >= mod) x -= mod;
}
long long xo(long long i) {
  if ((i & 3) == 0) return i;
  if ((i & 3) == 1) return 1;
  if ((i & 3) == 2) return i + 1;
  return 0;
}
map<long long, long long> help;
vector<long long> a;
int main() {
  long long n, i, j, k;
  cin >> n;
  long long a1, a2, diff = mod;
  for (long long i = 1; i <= n; i++) {
    if (n % i == 0) {
      if (abs(i - n / i) < diff) {
        a1 = min(i, n / i);
        a2 = max(i, n / i);
        diff = abs(i - n / i);
      }
    }
  }
  cout << a1 << " " << a2;
}

### 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;
template <class T>
inline void umax(T &a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void umin(T &a, T b) {
  if (a > b) a = b;
}
template <class T>
inline T abs(T a) {
  return a > 0 ? a : -a;
}
template <class T>
inline T gcd(T a, T b) {
  return __gcd(a, b);
}
template <class T>
inline T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
long long modpow(long long a, long long n, long long temp) {
  long long res = 1, y = a;
  while (n > 0) {
    if (n & 1) res = (res * y) % temp;
    y = (y * y) % temp;
    n /= 2;
  }
  return res % temp;
}
long long ison(long long mask, long long pos) { return (mask & (1 << pos)); }
long long cbit(long long n) {
  long long k = 0;
  while (n) n &= (n - 1), k++;
  return k;
}
long long nbit(long long n) {
  long long k = 0;
  while (n) n /= 2, k++;
  return k;
}
long long mod = 1e9 + 7;
int sgn(long long x) { return x < 0 ? -1 : !!x; }
const double eps = 1e-7;
const double pi = acos(-1.0);
inline void inc(long long &x, long long y) {
  x += y;
  if (x >= mod) x -= mod;
}
inline void dec(long long &x, long long y) {
  x -= y;
  if (x < 0) x += mod;
}
inline void chmax(long long &x, long long y) {
  if (y > x) x = y;
}
inline void mulm(long long &x, long long y) {
  x *= y;
  if (x >= mod) x -= mod;
}
long long xo(long long i) {
  if ((i & 3) == 0) return i;
  if ((i & 3) == 1) return 1;
  if ((i & 3) == 2) return i + 1;
  return 0;
}
map<long long, long long> help;
vector<long long> a;
int main() {
  long long n, i, j, k;
  cin >> n;
  long long a1, a2, diff = mod;
  for (long long i = 1; i <= n; i++) {
    if (n % i == 0) {
      if (abs(i - n / i) < diff) {
        a1 = min(i, n / i);
        a2 = max(i, n / i);
        diff = abs(i - n / i);
      }
    }
  }
  cout << a1 << " " << a2;
}

```