codefroces 723c Polycarp at the Radio

C. Polycarp at the Radio

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented as a sequence a1, a2, ..., an, where ai 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 bj 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, ..., bm will be as large as possible.

Find this maximum possible value of the minimum among the bj (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, ..., an (1 ≤ ai ≤ 109), where ai 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 bj (1 ≤ j ≤ m), where bj 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.

Examples

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.

题意：n个歌曲，m个乐队。给出每首歌是哪个乐队唱的即ai~an，通过修改歌曲的表演乐队，让你找到m个乐队中的每个乐队的最大可能表演数并且保证乐队（b1~bm）的值最大，和最小修改数，输出最后的歌曲演唱信息。

思路：直接模拟，记录下位置，然后依次修改，每个乐队的最大表演数那就是ave = n/m，把每个乐队的个数修改成ave，多的不用管，少的要补全

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>

using namespace std;

int n,m;
int a[2005];
vector<int>brand[2005];

int main() {
    cin >> n >> m;
    for(int i = 1; i <= n; i++) {
        cin >> a[i];
        if(a[i] > m) {
            brand[0].push_back(i);//比赛的时候没理解题意，写成了这样 if(a[i]>m)a[i] = 0;brand[a[i]].push_back(i)
        }else{
            brand[a[i]].push_back(i);
        }
    }//上述那样处理之后，修改的次数就会变多了比如输入//5 4  3 1 495987801 522279660 762868488 ；ave = 1;修改成3 1 2 4 762868488；这样的次数才是最少的，并不是把最后一个修改
    int ave = n/m;
    int tol = 0;
    for(int i = 1; i <= m; i++) {
        if(brand[i].size() < ave) {
            int change = ave-brand[i].size();
            tol += change;
            for(int j = 0; j <= m && change>0; j++) {
                vector<int>::iterator it = brand[j].begin();
                while(j == 0 && brand[j].size() > 0 && change>0) {
                    int pos = *it;
                    change--;
                    a[pos] = i;
                    brand[i].push_back(pos);
                    brand[j].erase(it);
                }
                while(j > 0 && brand[j].size() > ave && change>0) {
                    int pos = *it;
                    change--;
                    a[pos] = i;
                    brand[i].push_back(pos);
                    brand[j].erase(it);
                }
            }
        }
    }
    cout << ave << " " << tol << endl;
    cout << a[1];
    for(int i = 2; i <= n; i++) {
        cout << " " << a[i];
    }
    cout << endl;
    return 0;
}

PS：真的在玩耍？？？？？，题意理解错，说的是保证bi值最大就好了，很激动的修改了歌曲表演者中>m的值，服了自己的智商，读题真的很不认真，无话可说了！！！！