Salem gave you nn sticks with integer positive lengths a1,a2,…,ana1,a2,…,an.
For every stick, you can change its length to any other positive integer length (that is, either shrink or stretch it). The cost of changing the stick's length from aa to bb is |a−b||a−b|, where |x||x| means the absolute value of xx.
A stick length aiai is called almost good for some integer tt if |ai−t|≤1|ai−t|≤1.
Salem asks you to change the lengths of some sticks (possibly all or none), such that all sticks' lengths are almost good for some positive integer tt and the total cost of changing is minimum possible. The value of tt is not fixed in advance and you can choose it as any positive integer.
As an answer, print the value of tt and the minimum cost. If there are multiple optimal choices for tt, print any of them.
Input
The first line contains a single integer nn (1≤n≤10001≤n≤1000) — the number of sticks.
The second line contains nn integers aiai (1≤ai≤1001≤ai≤100) — the lengths of the sticks.
Output
Print the value of tt and the minimum possible cost. If there are multiple optimal choices for tt, print any of them.
Examples
Input
3 10 1 4
Output
3 7
Input
5 1 1 2 2 3
Output
2 0
Note
In the first example, we can change 11 into 22 and 1010 into 44 with cost |1−2|+|10−4|=1+6=7|1−2|+|10−4|=1+6=7 and the resulting lengths [2,4,4][2,4,4] are almost good for t=3t=3.
In the second example, the sticks lengths are already almost good for t=2t=2, so we don't have to do anything.
解法：枚举1-100的数，因为a[i]在1-100的范围内
代码：
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int a[1005];
int main()
{
int n;
cin>>n;
for(int t=0;t<n;t++)
{
scanf("%d",&a[t]);
}
int maxn=9999999;
int k;
for(int t=1;t<=100;t++)
{
int sum=0;
for(int j=0;j<n;j++)
{
sum+=min(fabs(a[j]-t+1),min(fabs(a[j]-t),fabs(a[j]-t-1)));
}
if(sum<maxn)
{
maxn=sum;
k=t;
}
}
cout<<k<<" "<<maxn<<endl;
return 0;
}