CodeForces - 439D Devu and his Brother 思维+排序

http://codeforces.com/problemset/problem/439/D

D. Devu and his Brother

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Devu and his brother love each other a lot. As they are super geeks, they only like to play with arrays. They are given two arrays a and bby their father. The array a is given to Devu and b to his brother.

As Devu is really a naughty kid, he wants the minimum value of his array a should be at least as much as the maximum value of his brother's array b.

Now you have to help Devu in achieving this condition. You can perform multiple operations on the arrays. In a single operation, you are allowed to decrease or increase any element of any of the arrays by 1. Note that you are allowed to apply the operation on any index of the array multiple times.

You need to find minimum number of operations required to satisfy Devu's condition so that the brothers can play peacefully without fighting.

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105). The second line will contain n space-separated integers representing content of the array a (1 ≤ ai ≤ 109). The third line will contain m space-separated integers representing content of the array b (1 ≤ bi ≤ 109).

Output

You need to output a single integer representing the minimum number of operations needed to satisfy Devu's condition.

Examples

input

2 2
2 3
3 5

output

3

input

3 2
1 2 3
3 4

output

4

input

3 2
4 5 6
1 2

output

0

Note

In example 1, you can increase a1 by 1 and decrease b2 by 1 and then again decrease b2 by 1. Now array a will be [3; 3] and array b will also be [3; 3]. Here minimum element of a is at least as large as maximum element of b. So minimum number of operations needed to satisfy Devu's condition are 3.

In example 3, you don't need to do any operation, Devu's condition is already satisfied.

题意：给出数列a，b。可以对每个数进行操作+1或者-1。求最小的操作数使得a数列的最小值，大于b数列的最大值。

题解：首先取出a数列的最小值x与b数列的最大值y进行操作(只是保证满足题意，取min(n,m)对x，y)。最少的操作肯定就是变到同一个数c(c是任意的)，那么有ans=(y-c)+(c-x)=y-x。这个最终的c介于这么多对的x，y之间。

代码：

#include
#define debug cout<<"aaa"<y;
}

int main(){
	int n,m;
	LL ans=0;
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	for(int i=1;i<=m;i++){
		cin>>b[i];
	}
	sort(a+1,a+n+1);
	sort(b+1,b+m+1,cmp);
	for(int i=1;i<=min(n,m);i++){
		if(a[i]