Codeforces Round #533 (Div. 2) A. Salem and Sticks（暴力）

A. Salem and Sticks

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

Copy

3
10 1 4

output

Copy

3 7

input

Copy

5
1 1 2 2 3

output

Copy

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.

题意：给你一个含有N个数的数组，让你找一个数x，使x的数组求值消耗的能量尽量小。

数组对于x的求值是把数组的每一个数削成| x-a[i] | <=1 ，并且a[i]数值变化1就要消耗1能量。

思路：

数据范围很小，直接暴力从1到max ( a[i] ) +2 试一试，找出那个最小的就行了，

细节看代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"==  "<<x<<"  =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
int n;
int a[maxn];
int ans=inf;
int cnt=inf;
int tt;
int tcnt;
int main()
{
    gbtb;
    cin>>n;
    repd(i,1,n)
    {
        cin>>a[i];
    }
    repd(i,1,2000)
    {
        tt=i;
        tcnt=0;
        repd(j,1,n)
        {
            if(abs(tt-a[j])<=1)
            {

            }else
            {
                tcnt+=abs(tt-a[j])-1;
            }
        }
        if(tcnt<cnt)
        {
            cnt=tcnt;
            ans=tt;
        }
    }
    cout<<ans<<" "<<cnt;
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}