【74.00%】【codeforces 747A】Display Size

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】:http://codeforces.com/contest/747/problem/A

【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 <bits/stdc++.h>
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<int,int> pii;
typedef pair<LL,LL> 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<<" "<<ansb<<endl;
    return 0;
}