#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 1e1 + 4;
vector<int> circle;
int parent[N];
int chk[N];
int color[N];
int tim[N];
int dis[N];
int position[N];
vector<int> adj[N];
vector<int> adj1[N];
vector<int> graph[N];
bool has_cycle;
int maxdis, maxnode, Totnode, depth = 1;
bool ok;
queue<int> q;
stack<int> stk;
vector<int> solution;
int indegree[N];
int go[N];
int to[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  int cas = 0;
  ll n, m, i, j, cnt = 0, cnt1 = 0, cnt2 = 0, even = 0, odd = 0, len, k, r, l,
                 z = 0, x = 0, y = 0, flag = 0, sum = 0;
  ll a = 0, b = 0, c = 0, d = 0, p, q, ans = 0, rem, quot, zero = 0, fst = 0,
     null = 0, snd = 0, lst = 0, rone = 0, one = 0, pos = 0, neg = 0,
     mn = INT_MAX, mx = INT_MIN;
  char ch;
  int h1, h2, m1, m2, h;
  int velo1, velo2, ac1, ac2, tim;
  int node, edge, u, v, cost;
  int best, worst;
  double nd, ad, bd, xd;
  string str, str1 = "", str2 = "", str3 = "", strstore1 = "", strstore2 = "";
  cin >> n;
  x = n;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      a = i;
      b = n / i;
      if ((b - a) < x) {
        x = b - a;
      }
    }
  }
  if (a == 0 && b == 0) {
    a = 1;
    b = n;
  }
  cout << a << " " << b << endl;
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
using ll = long long;
const int N = 1e1 + 4;
vector<int> circle;
int parent[N];
int chk[N];
int color[N];
int tim[N];
int dis[N];
int position[N];
vector<int> adj[N];
vector<int> adj1[N];
vector<int> graph[N];
bool has_cycle;
int maxdis, maxnode, Totnode, depth = 1;
bool ok;
queue<int> q;
stack<int> stk;
vector<int> solution;
int indegree[N];
int go[N];
int to[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int t;
  int cas = 0;
  ll n, m, i, j, cnt = 0, cnt1 = 0, cnt2 = 0, even = 0, odd = 0, len, k, r, l,
                 z = 0, x = 0, y = 0, flag = 0, sum = 0;
  ll a = 0, b = 0, c = 0, d = 0, p, q, ans = 0, rem, quot, zero = 0, fst = 0,
     null = 0, snd = 0, lst = 0, rone = 0, one = 0, pos = 0, neg = 0,
     mn = INT_MAX, mx = INT_MIN;
  char ch;
  int h1, h2, m1, m2, h;
  int velo1, velo2, ac1, ac2, tim;
  int node, edge, u, v, cost;
  int best, worst;
  double nd, ad, bd, xd;
  string str, str1 = "", str2 = "", str3 = "", strstore1 = "", strstore2 = "";
  cin >> n;
  x = n;
  for (i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      a = i;
      b = n / i;
      if ((b - a) < x) {
        x = b - a;
      }
    }
  }
  if (a == 0 && b == 0) {
    a = 1;
    b = n;
  }
  cout << a << " " << b << endl;
  return 0;
}

```