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:
import java.util.*;
import java.io.*;
import java.math.*;
import java.math.BigInteger;
import java.text.DecimalFormat;
public class Tester {
	public static int[] parent;
	static int arr[];;
	public static void main(String args[])
	{
		InputReader in = new InputReader(System.in);
		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);
		int n=in.nextInt(),root=-1;
		parent=new int[n];
		for(int i=0;i<n;i++)
		{
			parent[i]=i;
		}
		arr=new int[n];
		arr=nextIntArray(in,n);
		for(int i=0;i<n;i++)
		{
			parent[i]=i;
			if(arr[i]==i)
				root=i;
		}
		int count=0;
		for(int i=0;i<n;i++)
		{
			if(find(parent,i)!=find(parent,arr[i]))
			{
				merge(i,arr[i]);
			}
			else
			{
				if(root==-1)
				{
					root=find(parent,i);
				}
				if(arr[i]!=root)
				{
					arr[i]=root;
					count++;
				}
			}
		}
		out.println(count);
		for(int i=0;i<n;i++)
		{
			out.print((arr[i]+1)+" ");
		}
		out.close();
	}
	static int find(int[] arr,int x)
	{
		if(arr[x]==x)
			return x;
		return find(arr,parent[x]);
	}
	static void merge(int x,int y)
	{
		int px=find(parent,x);
		int py=find(parent,y);
		parent[px]=py;
	}
	private static int[] nextIntArray(InputReader in,int n)	
	{
		int[] a=new int[n];
		for(int i=0;i<n;i++)
			a[i]=in.nextInt()-1;
		return a;
	}
	public static double dis(double x1,double x2,double y1)
	{
		return Math.sqrt(((x1-x2)*(x1-x2))+(y1*y1));
	}
	private static String[] nextStringArray(InputReader in,int n)	
	{
		String[] a=new String[n];
		for(int i=0;i<n;i++)
			a[i]=in.next();
		return a;
	}
	private static void show(int[] a)
	{
		for(int i=0;i<a.length;i++)
			System.out.print(a[i]+" ");
		System.out.println();
	}
	private static void show2DArray(char[][] a)
	{
		for(int i=0;i<a.length;i++)
		{
			for(int j=0;j<a[0].length;j++)
				System.out.print(a[i][j]);
			System.out.println();
		}
	}
 static long fact(long x)
 {
	 long i=x,mul=1;
	 while(i>0)
	 {
		 mul=(mul%1000000007)*(i%1000000007)%1000000007;
		 i--;
	 }
	 return mul;
 }
 static class LengthComparator implements Comparator<String> {
	    public int compare(String arg0, String arg1) {
		// Use Integer.compare to compare the two Strings' lengths.
		return (arg0+arg1).compareTo(arg1+arg0);
	    }
	}
// static long output(ArrayList<Integer> h[],int j,boolean[] v)
// {
//	 int k;
//	 v[j]=true;
//	 sum++;
//	 for(k=0;k<h[j].size();k++)
//	 {
//		 if(v[h[j].get(k)]==false)
//		 {
//			 output(h,h[j].get(k),v);
//		 }
//	 }
//	 return sum;
// }
 static long func(boolean v[],int j,ArrayList<Integer> h[],long ban[])
 {
	 	v[j]=true;
	 	 int k;
	 	long sum=0;
	    sum=sum+ban[j];
	 	//System.out.println(h[j].size());
	 	for(k=0;k<h[j].size();k++)
	 	{
	 		if(v[h[j].get(k)]==false)
	 		{
	 			v[h[j].get(k)]=true;
	 			sum+=func(v,h[j].get(k),h,ban);
	 		}
	 	}
	 	return sum;
 }
/*static class Graph {
    	
    	private static Deque<Integer> stack = new ArrayDeque<Integer>();
 		private int least,count,v;
 		
		Set<Integer>[] cities;
    	private int[] risk;
    	
    	Graph(int n,String[] risk){
    		
    		cities = new HashSet[n];
    		this.risk = new int[n];
    		for(int i =0;i<n;i++){
    			cities[i] = new HashSet<>();
    		}
    		for(int i =0;i<n;i++){
    			this.risk[i] = Integer.parseInt(risk[i]);
    		}
    		visited = new boolean[n];
    	}
    	
    	
    	public void add(int x,int y){
    		cities[x].add(y);
    		cities[y].add(x);
    	}
}*/
	static int root(int arr[],int i)
	{
		while(arr[i]!=i)
		{
			i=arr[i];
		}
		return i;
	}
	static boolean find(int arr[],int a,int b)
	{
		if(root(arr,a)==root(arr,b))
		{
			return true;
		}
		else
			return false;
	}
	static void weighted_union(int arr[],int size[],int a,int b)
	{
		int root_a=root(arr,a);
		int root_b=root(arr,b);
		if(root_a!=root_b)
		{
			if(size[root_a]<size[root_b])
			{
				arr[root_a]=arr[root_b];
				size[root_b]+=size[root_a];
			}
			else
			{
				arr[root_b]=arr[root_a];
				size[root_a]+=size[root_b];
			}
		//	count--;
		}
	}
	static class Pair implements Comparable<Pair>
	{
		private long first;
		private long index;
		//private long second;;

		public Pair(long i, long j) 
		{ 
			this.first = i;
			this.index = j;
		}
		public long getFirst() { return first; }
		//public long getSecond() { return second; }
		public long getIndex() { return index ;}
		public void setFirst(long k) { this.first=k ; }
		public void setIndex(long k) { this.index=k ;}
		//public void setSecond(long k) { this.second=k ;}
		@Override
		public int compareTo(Pair o) 
		{
			return Long.compare(this.first, o.first);
		}
	}
	static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream inputstream) {
			reader = new BufferedReader(new InputStreamReader(inputstream));
			tokenizer = null;
		}

		public String nextLine(){
			String fullLine=null;
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					fullLine=reader.readLine();
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
				return fullLine;
			}
			return fullLine;
		}
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
	}

}