CodeForces723C Polycarp at the Radio 贪心

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.

贪心题，最大值为 n/m，然后最少的操作数是将所有次数小于n/m的数的次数变为n/m所需花的操作次数之和。然后需要改动的数有两种可能，一种是数值大于m，另一种是数值小于等于吗，但出现次数大于n/m。

#include <iostream>
#include <cstdio>
#include <map>
#include <set>
#include <vector>
#include <queue>
#include <stack>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
#define INF 0x3f3f3f3f
typedef long long LL;
double PI=acos(-1);
int n,m,x;
int a[2005],b[2005];

int check(int x)
{
    if(b[x]>=n/m){
        while(b[x]>=n/m){
            x++;
        }
    }
    return x;
}
int main()
{
    scanf("%d%d",&n,&m);
    memset(b,0,sizeof(b));
    int ans=n/m;
    for(int i=0;i<n;i++){
        scanf("%d",a+i);
        if(a[i]<=m){
            b[a[i]]++;
        }
    }
    int cnt=0;
    for(int i=1;i<=m;i++){
        if(b[i]<ans){
            cnt+=ans-b[i];
        }
    }
    x=check(1);
    printf("%d %d\n",ans,cnt);
    for(int i=0;i<n&&cnt;i++){
        if(a[i]>m||a[i]<=m&&b[a[i]]>ans){
            if(a[i]<=m&&b[a[i]]>ans){
                b[a[i]]--;
            }
            a[i]=x;
            b[x]++;
            x=check(x);
            cnt--;
        }
    }
    for(int i=0;i<n;i++){
        printf(i==n-1?"%d\n":"%d ",a[i]);
    }
    return 0;
}