Problem: A tree is an undirected connected graph without cycles.

Let's consider a rooted undirected tree with n vertices, numbered 1 through n. There are many ways to represent such a tree. One way is to create an array with n integers p1, p2, ..., pn, where pi denotes a parent of vertex i (here, for convenience a root is considered its own parent).

<image> For this rooted tree the array p is [2, 3, 3, 2].

Given a sequence p1, p2, ..., pn, one is able to restore a tree:

  1. There must be exactly one index r that pr = r. A vertex r is a root of the tree. 
  2. For all other n - 1 vertices i, there is an edge between vertex i and vertex pi. 



A sequence p1, p2, ..., pn is called valid if the described procedure generates some (any) rooted tree. For example, for n = 3 sequences (1,2,2), (2,3,1) and (2,1,3) are not valid.

You are given a sequence a1, a2, ..., an, not necessarily valid. Your task is to change the minimum number of elements, in order to get a valid sequence. Print the minimum number of changes and an example of a valid sequence after that number of changes. If there are many valid sequences achievable in the minimum number of changes, print any of them.

Input

The first line of the input contains an integer n (2 ≤ n ≤ 200 000) — the number of vertices in the tree.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ n).

Output

In the first line print the minimum number of elements to change, in order to get a valid sequence.

In the second line, print any valid sequence possible to get from (a1, a2, ..., an) in the minimum number of changes. If there are many such sequences, any of them will be accepted.

Examples

Input

4
2 3 3 4


Output

1
2 3 4 4 


Input

5
3 2 2 5 3


Output

0
3 2 2 5 3 


Input

8
2 3 5 4 1 6 6 7


Output

2
2 3 7 8 1 6 6 7

Note

In the first sample, it's enough to change one element. In the provided output, a sequence represents a tree rooted in a vertex 4 (because p4 = 4), which you can see on the left drawing below. One of other correct solutions would be a sequence 2 3 3 2, representing a tree rooted in vertex 3 (right drawing below). On both drawings, roots are painted red.

<image>

In the second sample, the given sequence is already valid.

Solution (in JAVA):

Solution:
/* My Approach :
    Make a graph :
     case 1 : if there is only one connected component 
              1) if a node exist such that p[i]=i then do nothing
              2) else find a vertex which is present in the cycle make it p[i]=i and ans=1

     case 2 : if there are more than one connected component
              every connectec omponenet has a cycle . it may be of length 1
             1) if at least one connected component exist such that p[i]=i;
                then ans=cc-1; and from every connected component pick a vertex v whihc is presnet in cycle and make p[v]=i;
             2) else ans=cc . becaue we have to change in every connectec comp0nent.
                pick a vertex from a cc which is present in cycle and make p[i]=i;
                and for other component pick a cyclic vertex and make p[v]=i; 
*/
import java.util.*;
import java.io.*;
public class c{ 

static PrintWriter out = new PrintWriter(System.out);
static int flag=0;
static void dfs(LinkedList<Integer> l[],int s,int p,boolean v[],HashSet<Integer> h){
                       v[s]=true;
                     for(Integer i: l[s])
                          if(v[i]==false)
                             dfs(l,i,s,v,h);
                          else if(i!=p&&flag==0)
                            { h.add(i); flag=1;}

}
public static void main(String[] args)throws IOException { 
 

        int n = ni();
        int p[] = new int[n];
        LinkedList<Integer> l[] = new LinkedList[n];
               for(int i=0;i<n;i++) l[i] = new LinkedList();
          int f=-1;
         for(int i=0;i<n;i++) {
                   p[i] = ni()-1;
                    if(p[i]==i)f=i;
                   l[i].add(p[i]);
                   l[p[i]].add(i);
          }
        boolean v[] = new boolean[n];
        HashSet<Integer> h = new HashSet<Integer>();
        int cc=0,ans=0;
           for(int i=0;i<n;i++)
                if(v[i]==false)
                  { cc++;flag=0; dfs(l,i,-1,v,h);}
          if(cc!=1){
                      if(f==-1){ 
                                  for(Integer i: h){
                                        if(f==-1){
                                            f=i;   
                                            p[i]=i;
                                         }
                                        else
                                          p[i]=f;
                                   }
                          ans=cc;
                      }
                    else{
                            for(Integer i: h)
                                 if(i!=f)
                                    p[i]=f;   
                           ans=cc-1;             
                     }
          }
        else if(f==-1){
                 for(Integer i : h)
                      p[i]=i;
                      ans =1;
          }
             out.println(ans);
          for(int i=0;i<n;i++)
             out.print((p[i]+1)+" ");
      out.flush();
}   
    static FastReader sc=new FastReader(); 
 
      static int ni(){
                 int x = sc.nextInt();
                 return(x);
        }
      static long nl(){
              long x = sc.nextLong();
              return(x);
         }
      static String n(){
                 String str = sc.next();
                     return(str);
       }
     static String ns(){
                 String str = sc.nextLine();
                   return(str);
      }
     static double nd(){
               double d = sc.nextDouble();
                 return(d);
       }
  static class FastReader 
    { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastReader() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    } 
}