cf-723C Polycarp at the Radio

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于 2018-01-28 11:23:46 发布

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本文介绍了一个播放列表优化问题，目标是最小化热门乐队表演歌曲数量间的差距，并减少播放列表的改动次数。通过计算平均值和调整乐队表演频次，实现播放列表的最优配置。

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cf-723C Polycarp at the Radio

题目点击打开链接 http://codeforces.com/contest/723/problem/C

分析：先定义每个数字在数组中出现次数叫做这个数的频率，注意是次数。

题意：给出两个数字 n 和 m，再给出 n 个数字放在数组 a[i] 里。使数组 a 里面的数全部变为 1~m 之间的数字，并且使每个数的频率尽量大，也就是每个数的频率高于平均值。然后输出数组改变后所有数频率中的最小值和改变数组时改变了几次，以及改变后的数组。

可以知道平均值就是 n/m，然后在数组a中找到符合 1~m 的数字并记录下次数，如果高于平均值就再找下一个，低于平均值则记下差几个数能够到达平均值，这样可以求出改变次数。最后在数组中找到所有不符合条件的数【频率太小需要将频率大的改为这个数，大于m的需要变小】改掉

Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented as a sequence a₁, a₂, ..., a_n, where a_i is a band, which performs the i-th song. Polycarp likes bands with the numbers from 1 to m, but he doesn't really like others.

We define as b_j the number of songs the group j is going to perform tomorrow. Polycarp wants to change the playlist in such a way that the minimum among the numbers b₁, b₂, ..., b_m will be as large as possible.

Find this maximum possible value of the minimum among the b_j (1 ≤ j ≤ m), and the minimum number of changes in the playlist Polycarp needs to make to achieve it. One change in the playlist is a replacement of the performer of the i-th song with any other group.

Input

The first line of the input contains two integers n and m (1 ≤ m ≤ n ≤ 2000).

The second line contains n integers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹), where a_i is the performer of the i-th song.

Output

In the first line print two integers: the maximum possible value of the minimum among the b_j (1 ≤ j ≤ m), where b_j is the number of songs in the changed playlist performed by the j-th band, and the minimum number of changes in the playlist Polycarp needs to make.

In the second line print the changed playlist.

If there are multiple answers, print any of them.

Example

Input

4 2
1 2 3 2

Output

2 1
1 2 1 2

Input

7 3
1 3 2 2 2 2 1

Output

2 1
1 3 3 2 2 2 1

Input

4 4
1000000000 100 7 1000000000

Output

1 4
1 2 3 4

Note

In the first sample, after Polycarp's changes the first band performs two songs (b₁ = 2), and the second band also performs two songs (b₂ = 2). Thus, the minimum of these values equals to 2. It is impossible to achieve a higher minimum value by any changes in the playlist.

In the second sample, after Polycarp's changes the first band performs two songs (b₁ = 2), the second band performs three songs (b₂ = 3), and the third band also performs two songs (b₃ = 2). Thus, the best minimum value is 2.

#include<iostream>
#include<cstdio>
#include<cctype>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=2050;
int a[N],val[N],p[N];
int pos[2005];
int main(){
	int n,m,min_ave,min_cnt;
	memset(pos,0,sizeof(pos));
	cin>>n>>m;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	min_ave=n/m;
	min_cnt=0;
	for(int i=1;i<=m;i++){
		int num=0;
		for(int j=1;j<=n;j++){
			if(a[j]==i){
				pos[j]=1;
				num++;
			}
			if(num>=min_ave){
				break;
			}
		}
		if(num<min_ave){
			val[i]=min_ave-num;
			min_cnt+=val[i];
		}
	}
	int count=1;
	for(int i=1;i<=m;i++){
		for(int j=0;j<val[i];j++){
			while(pos[count]==1){
				count++;
			}
			a[count++]=i;
		}
	}
	cout<<min_ave<<" "<<min_cnt<<endl;
	for(int i=1;i<=n;i++){
		cout<<a[i]<<" ";
	}
	cout<<endl;
	return 0;
}