#include <bits/stdc++.h>
using namespace std;
string getbin(int n) {
  string ans = "";
  while (n) {
    ans.push_back(n % 2 + '0');
    n /= 2;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
int getbit(int n, int i) { return ((n & (1 << i)) != 0); }
int setbit(int n, int i) { return ((n | (1 << i))); }
int clearbit(int n, int i) { return (n & (~(1 << i))); }
int updatebit(int n, int i, int v) {
  int mask = ~(1 << i);
  int value = (v << i);
  return ((n & mask) | value);
}
int insert(int n, int m, int end, int start) {
  int num = 0;
  int base = 1;
  for (int i = 0; i < start; ++i) {
    num += ((n & (1 << i)) != 0) * base;
    base *= 2;
  }
  int mask = (m << start);
  num = (-1 << (end + 1)) | num;
  n &= num;
  n |= mask;
  return n;
}
long long pow(long long n, long long a) {
  long long ans = 1;
  for (long long i = 1; i <= a; ++i) {
    ans *= n;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int n1 = 0, n2 = 0;
  vector<int> v;
  int minn = INT_MAX;
  for (int i = 1; i <= n; ++i) {
    if (n % i == 0) {
      v.push_back(i);
    }
  }
  for (int i = 0, j = v.size() - 1; i <= j; ++i, --j) {
    if (v[i] * v[j] == n && v[j] - v[i] < minn) {
      n1 = v[i];
      n2 = v[j];
      minn = n2 - n1;
    }
  }
  cout << n1 << " " << n2 << endl;
}

### Prompt

Create a solution in CPP for 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;
string getbin(int n) {
  string ans = "";
  while (n) {
    ans.push_back(n % 2 + '0');
    n /= 2;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
int getbit(int n, int i) { return ((n & (1 << i)) != 0); }
int setbit(int n, int i) { return ((n | (1 << i))); }
int clearbit(int n, int i) { return (n & (~(1 << i))); }
int updatebit(int n, int i, int v) {
  int mask = ~(1 << i);
  int value = (v << i);
  return ((n & mask) | value);
}
int insert(int n, int m, int end, int start) {
  int num = 0;
  int base = 1;
  for (int i = 0; i < start; ++i) {
    num += ((n & (1 << i)) != 0) * base;
    base *= 2;
  }
  int mask = (m << start);
  num = (-1 << (end + 1)) | num;
  n &= num;
  n |= mask;
  return n;
}
long long pow(long long n, long long a) {
  long long ans = 1;
  for (long long i = 1; i <= a; ++i) {
    ans *= n;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  int n1 = 0, n2 = 0;
  vector<int> v;
  int minn = INT_MAX;
  for (int i = 1; i <= n; ++i) {
    if (n % i == 0) {
      v.push_back(i);
    }
  }
  for (int i = 0, j = v.size() - 1; i <= j; ++i, --j) {
    if (v[i] * v[j] == n && v[j] - v[i] < minn) {
      n1 = v[i];
      n2 = v[j];
      minn = n2 - n1;
    }
  }
  cout << n1 << " " << n2 << endl;
}

```