D2. Sage‘s Birthday (hard version)

D2. Sage's Birthday (hard version)

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

This is the hard version of the problem. The difference between the versions is that in the easy version all prices aiai are different. You can make hacks if and only if you solved both versions of the problem.

Today is Sage's birthday, and she will go shopping to buy ice spheres. All nn ice spheres are placed in a row and they are numbered from 11 to nn from left to right. Each ice sphere has a positive integer price. In this version, some prices can be equal.

An ice sphere is cheap if it costs strictly less than two neighboring ice spheres: the nearest to the left and the nearest to the right. The leftmost and the rightmost ice spheres are not cheap. Sage will choose all cheap ice spheres and then buy only them.

You can visit the shop before Sage and reorder the ice spheres as you wish. Find out the maximum number of ice spheres that Sage can buy, and show how the ice spheres should be reordered.

Input

The first line contains a single integer nn (1≤n≤105)(1≤n≤105) — the number of ice spheres in the shop.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤109)(1≤ai≤109) — the prices of ice spheres.

Output

In the first line print the maximum number of ice spheres that Sage can buy.

In the second line print the prices of ice spheres in the optimal order. If there are several correct answers, you can print any of them.

Example

input

Copy

7
1 3 2 2 4 5 4

output

Copy

3
3 1 4 2 4 2 5 

Note

In the sample it's not possible to place the ice spheres in any order so that Sage would buy 44 of them. If the spheres are placed in the order (3,1,4,2,4,2,5)(3,1,4,2,4,2,5), then Sage will buy one sphere for 11 and two spheres for 22 each.

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最尴尬的不是D1会做而D2不会做，而是用可以过D2的方法先做了D1，完事之后不敢用这个代码再去交D2.....

其实这两个题完全是一个题目，就是一个基本的贪心....不明白为什么分得分成两种情况

#include
# include
# include
 
using namespace std;
typedef long long int  ll;
int a[100000+10];
int ans[100000+10];
int main ()
{
 
    int n;
    cin>>n;
 
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    sort(a+1,a+1+n);
    int len=1;
    for(int i=2;i<=n;i+=2)
    {
        ans[i]=a[len];
        len++;
    }
    for(int i=1;i<=n;i+=2)
    {
        ans[i]=a[len];
        len++;
    }
    int cnt=0;
    for(int i=2;i
    {
        if(ans[i]-1]&&ans[i]1])
            cnt++;
    }
    cout<
    for(int i=1;i<=n;i++)
    {
        cout<" ";
    }
    return 0;
 
}