codeforces 723C : Polycarp at the Radio

Description

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.

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.

 

正解：贪心

解题报告：

　　最优的次数显然，方案的话，每次暴力找到一个多余的，并把多出来的部分给别的缺少的即可。

//It is made by jump~
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <ctime>
#include <vector>
#include <queue>
#include <map>
#include <set>
using namespace std;
typedef long long LL;
const int inf = (1<<30);
const int MAXN = 2011;
int n,m,ans,zong;
int a[MAXN],cnt[MAXN];
bool pd[MAXN];

inline int getint()
{
    int w=0,q=0; char c=getchar();
    while((c<'0' || c>'9') && c!='-') c=getchar(); if(c=='-') q=1,c=getchar(); 
    while (c>='0' && c<='9') w=w*10+c-'0', c=getchar(); return q ? -w : w;
}

inline void work(){
    n=getint(); m=getint(); for(int i=1;i<=n;i++) a[i]=getint();
    for(int i=1;i<=n;i++) if(a[i]<=m) cnt[a[i]]++;
    ans=n/m; printf("%d ",ans); bool flag=false;
    for(int i=1;i<=n;i++) {
    if(a[i]<=m && cnt[a[i]]>ans) {
        flag=false;        
        for(int j=1;j<=m;j++) {
        if(cnt[j]<ans) {
            flag=true;
            cnt[j]++; cnt[a[i]]--;a[i]=j; break;
        }
        }
        if(flag) zong++;
    }
    else if(a[i]>m) {
        flag=false;        
        for(int j=1;j<=m;j++) {
        if(cnt[j]<ans) {
            flag=true;
            cnt[j]++; a[i]=j;
            break;
        }
        }
        if(flag) zong++;
    }
    }
    printf("%d
",zong);
    for(int i=1;i<=n;i++) printf("%d ",a[i]);
}

int main()
{
    work();
    return 0;
}