Polycarp is a music editor at the radio station. He received a playlist for tomorrow, that can be represented as a sequencea1, 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.
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.
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.
4 2 1 2 3 2
2 1 1 2 1 2
7 3 1 3 2 2 2 2 1
2 1 1 3 3 2 2 2 1
4 4 1000000000 100 7 1000000000
1 4 1 2 3 4
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 <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PI acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int n,m;
int flag[2005];
int num[2005];
int add[2005];
ll a[2005];
int main()
{
int maxx,re=0;
int len =0;
memset(a,0,sizeof(a));
memset(num,0,sizeof(num));
memset(add,0,sizeof(add));
memset(flag,0,sizeof(flag));
scanf("%d %d",&n,&m);
maxx = n/m;
for(int i=1;i<=n;i++)
{
scanf("%I64d",&a[i]);
if(a[i]<=m)
{
num[a[i]]++;
if(num[a[i]]>maxx)
flag[len++] = i;
}
else
flag[len++] = i;
}
for(int i=1;i<=m;i++)
{
if(num[i]<maxx)
{
add[i] = maxx-num[i];
re += add[i];
}
}
len = 0;
for(int i=1;i<=m;i++)
{
for(int j=1;j<=add[i];j++)
{
a[flag[len++]] = i;
}
}
printf("%d %d\n",maxx,re);
for(int i=1;i<=n;i++)
{
printf("%I64d",a[i]);
if(i!=n)
printf(" ");
}
printf("\n");
return 0;
}