#include <bits/stdc++.h>
using namespace std;
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename A, typename B>
istream &operator>>(istream &cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
const long double pi = 3.14159265358979323846;
const long long mod = 1000000007;
long long n, m, k, q, l, r, x, y, z;
const long long template_array_size = 1e6 + 1178;
long long a[template_array_size];
long long b[template_array_size];
long long c[template_array_size];
string second, t;
long long ans = 0;
void solve(int tc = 0) {
  pair<long long, long long> ans = make_pair(1e6, 1e6);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    if (n % i == 0) {
      ans = min(ans, make_pair(abs(n / i - i), min(n / i, i)));
    }
  }
  cout << ans.second << " " << ans.first + ans.second << '\n';
}
int main() {
  { ios_base::sync_with_stdio(false); }
  {
    cin.tie(NULL);
    cout.tie(NULL);
  }
  int tc = 1;
  for (int t = 0; t < tc; t++) solve(t);
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v);
template <typename A, typename B>
ostream &operator<<(ostream &cout, pair<A, B> const &p) {
  return cout << "(" << p.first << ", " << p.second << ")";
}
template <typename A>
ostream &operator<<(ostream &cout, vector<A> const &v) {
  cout << "[";
  for (int i = 0; i < v.size(); i++) {
    if (i) cout << ", ";
    cout << v[i];
  }
  return cout << "]";
}
template <typename A, typename B>
istream &operator>>(istream &cin, pair<A, B> &p) {
  cin >> p.first;
  return cin >> p.second;
}
mt19937 rng(std::chrono::steady_clock::now().time_since_epoch().count());
void usaco(string filename) {
  freopen((filename + ".in").c_str(), "r", stdin);
  freopen((filename + ".out").c_str(), "w", stdout);
}
const long double pi = 3.14159265358979323846;
const long long mod = 1000000007;
long long n, m, k, q, l, r, x, y, z;
const long long template_array_size = 1e6 + 1178;
long long a[template_array_size];
long long b[template_array_size];
long long c[template_array_size];
string second, t;
long long ans = 0;
void solve(int tc = 0) {
  pair<long long, long long> ans = make_pair(1e6, 1e6);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    if (n % i == 0) {
      ans = min(ans, make_pair(abs(n / i - i), min(n / i, i)));
    }
  }
  cout << ans.second << " " << ans.first + ans.second << '\n';
}
int main() {
  { ios_base::sync_with_stdio(false); }
  {
    cin.tie(NULL);
    cout.tie(NULL);
  }
  int tc = 1;
  for (int t = 0; t < tc; t++) solve(t);
}

```