time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

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.

【题目链接】:

【题解】

暴力枚举n的因子x

看看n/x是不是满足x<=n/x

然后记录那个差最小的就好.

【完整代码】

#include 
     
      using namespace std;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define LL long long#define rep1(i,a,b) for (int i = a;i <= b;i++)#define rep2(i,a,b) for (int i = a;i >= b;i--)#define mp make_pair#define pb push_back#define fi first#define se second#define rei(x) scanf("%d",&x)#define rel(x) scanf("%I64d",&x)typedef pair
      
        pii;typedef pair
       
         pll;//const int MAXN = x;const int dx[9] = {
    0,1,-1,0,0,-1,-1,1,1};const int dy[9] = {
    0,0,0,-1,1,-1,1,-1,1};const double pi = acos(-1.0);int n;int ma = 1e7,ansa,ansb;int main(){    //freopen("F:\\rush.txt","r",stdin);    rei(n);    rep1(i,1,n)    if ((n%i)==0)    {        int a = n/i;        int b = i;        if (a <= b)        {            if (b-a<=ma)            {                ma = b-a;                ansa = a;                ansb = b;            }        }    }    cout << ansa<<" "<
        
         <