C. Laboratory Work

time limit per test : 1 second

memory limit per test : 256 megabytes

input : standard input

output : standard output

Anya and Kirill are doing a physics laboratory work. In one of the tasks they have to measure some value \(n\) times, and then compute the average value to lower the error.

Kirill has already made his measurements, and has got the following integer values: \(x_1, x_2, …, x_n\). It is important that the values are close to each other, namely, the difference between the maximum value and the minimum value is at most \(2\).

• Anya does not want to make the measurements, however, she can't just copy the values from Kirill's work, because the error of each measurement is a random value, and this coincidence will be noted by the teacher. Anya wants to write such integer values \(y_1, y_2, …, y_n\) in her work, that the following conditions are met:

• the average value of \(x_1, x_2, …, x_n\) is equal to the average value of \(y_1, y_2, …, y_n\);

  all Anya's measurements are in the same bounds as all Kirill's measurements, that is, the maximum value among Anya's values is not greater than the maximum value among Kirill's values, and the minimum value among Anya's values is not less than the minimum value among Kirill's values;

• the number of equal measurements in Anya's work and Kirill's work is as small as possible among options with the previous conditions met. Formally, the teacher goes through all Anya's values one by one, if there is equal value in Kirill's work and it is not strike off yet, he strikes off this Anya's value and one of equal values in Kirill's work. The number of equal measurements is then the total number of strike off values in Anya's work.

Help Anya to write such a set of measurements that the conditions above are met.

Input

The first line contains a single integer \(n (1 ≤ n ≤ 100 000)\) — the numeber of measurements made by Kirill.

The second line contains a sequence of integers \(x_1, x_2, …, x_n ( - 100 000 ≤ x_i ≤ 100 000)\) — the measurements made by Kirill. It is guaranteed that the difference between the maximum and minimum values among values \(x_1, x_2, …, x_n\) does not exceed \(2\).

Output

In the first line print the minimum possible number of equal measurements.

In the second line print n integers \(y_1, y_2, …, y_n\) — the values Anya should write. You can print the integers in arbitrary order. Keep in mind that the minimum value among Anya's values should be not less that the minimum among Kirill's values, and the maximum among Anya's values should be not greater than the maximum among Kirill's values.

If there are multiple answers, print any of them.

Examples

input

6

-1 1 1 0 0 -1

output

2

0 0 0 0 0 0

input

3

100 100 101

output

3

101 100 100

input

7

-10 -9 -10 -8 -10 -9 -9

output

5

-10 -10 -9 -9 -9 -9 -9

Note

In the first example Anya can write zeros as here measurements results. The average value is then equal to the average value of Kirill's values, and there are only two equal measurements.

In the second example Anya should write two values \(100\) and one value \(101\) (in any order), because it is the only possibility to make the average be the equal to the average of Kirill's values. Thus, all three measurements are equal.

In the third example the number of equal measurements is \(5\).

题意

给出一个数组\(a\)，要求构造一个新的数组\(b\)，满足：

1. \(b\)数组的平均值和\(a\)数组的平均值相等，
2. \(b\)数组的最大值不大于\(a\)数组的最大值，最小值不小于\(a\)数组的最小值
3. 要求\(a,b\)数组不同的元素尽可能多

思路

因为题目中给出了：数组中最大值和最小值的差值不超过\(2\)，可以分成两种情况：

1. 当\(a\)数组中的最大值和最小值差值小于\(2\)的时候，无法构造出一个和\(a\)不同的满足条件的\(b\)数组，相同元素个数为\(n\)
2. 当最大值和最小值差值等于\(2\)时，可以将\(a\)数组转换成只有\(0,1,2\)的一个新数组，然后有两种方案进行构造：

• 方案一：将两个\(1\)变成一个\(0\)和一个\(2\)
• 方案二：将一个\(0\)和一个\(2\)变成两个\(1\)

当\(1\)较少的时候，选择方案二，当\(1\)比\(2\)和\(0\)的个数多的时候，选择方案一

代码

#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define ms(a,b) memset(a,b,sizeof(a))
const int inf=0x3f3f3f3f;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int maxn=1e6+10;
const int mod=1e9+7;
const int maxm=1e3+10;
using namespace std;
int a[maxn];
int num[maxn];
bool cmp(int a,int b)
{
    return a>b;
}
int main(int argc, char const *argv[])
{
    #ifndef ONLINE_JUDGE
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w", stdout);
        srand((unsigned int)time(NULL));
    #endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin>>n;
    int minn=inf;
    int maxx=-inf;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        minn=min(a[i],minn);
        maxx=max(maxx,a[i]);
    }
    if(maxx-minn<=1)
    {
        cout<<n<<endl;
        sort(a,a+n,cmp);
        for(int i=0;i<n;i++)
            cout<<a[i]<<" ";
        cout<<endl;
    }
    else
    {
        for(int i=0;i<n;i++)
            a[i]-=minn,num[a[i]]++;
        // 两个1变成一个0和一个2
        // 一个0和一个2变成两个1
        int res=min(num[0],num[2]);
        int res1=num[1]/2;
        // 如果1比较少，选择方案二
        if(res>res1)
        {
            num[1]+=res*2;
            num[0]-=res;
            num[2]-=res;
        }
        // 如果1比较多，选择方案一
        else
        {
            res=res1;
            num[1]-=res*2;
            num[0]+=res;
            num[2]+=res;
        }
        cout<<n-res*2<<endl;
        for(int i=0;i<3;i++)
            for(int j=0;j<num[i];j++)
                cout<<minn+i<<" ";
        cout<<endl;
    }
    #ifndef ONLINE_JUDGE
        cerr<<"Time elapsed: "<<1.0*clock()/CLOCKS_PER_SEC<<" s.\n";
    #endif
    return 0;
}