题干：

One way to create a task is to learn from math. You can generate some random math statement or modify some theorems to get something new and build a new task from that.

For example, there is a statement called the "Goldbach's conjecture". It says: "each even number no less than four can be expressed as the sum of two primes". Let's modify it. How about a statement like that: "each integer no less than 12 can be expressed as the sum of two composite numbers." Not like the Goldbach's conjecture, I can prove this theorem.

You are given an integer n no less than 12, express it as a sum of two composite numbers.

Input

The only line contains an integer n (12 ≤ n ≤ 106).

Output

Output two composite integers x and y (1 < x, y < n) such that x + y = n. If there are multiple solutions, you can output any of them.

Examples

Input

12

Output

4 8

Input

15

Output

6 9

Input

23

Output

8 15

Input

1000000

Output

500000 500000

Note

In the first example, 12 = 4 + 8 and both 4, 8 are composite numbers. You can output "6 6" or "8 4" as well.

In the second example, 15 = 6 + 9. Note that you can't output "1 14" because 1 is not a composite number.

题目大意：

  给你一个不小于12的整数，让你分解成两个合数。

解题报告：

  看似不显然，，但是其实很显然啊，如果是偶数那就输出4和n-4  或者6 和n-6这样的都行。如果是奇数那就输出9 和 n-9。做题的时候傻了吧唧的写了个素数打表。得亏数据给的小是1e6，，要是1e18这样的还真不好办了（不过要是给那样的数据也就回去想规律了2333）。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e6 + 5;
bool is[MAX];
void prime() {
	memset(is,1,sizeof is);
	is[0]=is[1]=0;
	for(int i = 2; i<=1000000; i++) {
		if(is[i]) {
			for(int j = i+i; j<=1000000; j+=i) is[j]=0; 
		}
	}
}
int main()
{
	prime();
	int n;
	cin>>n;
	for(int i = 3; i*2<=n; i++) {
		if(is[i]==0 && is[n-i] == 0) {
			printf("%d %d\n",i,n-i);break;
		}
	}

	return 0 ;
 }