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.io.*;
import java.util.Arrays;
import java.util.Random;

public class ProblemD {
  
  public static void main(String[] args) throws IOException {
    init();
    new ProblemD().run();
    out.flush();
    out.close();
  }
  
  static void init() throws IOException {
    in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    out = new PrintWriter(new OutputStreamWriter(System.out));
    in.ordinaryChars(0, 65535);
    in.wordChars(0, 65535);
    in.whitespaceChars(' ', ' ');
    in.whitespaceChars('\n', '\n');
    in.whitespaceChars('\r', '\r');
  }
  
  // класс для пар
  class Pair<L, R> {
    private final L X;
    private final R Y;
    
    public Pair(L X, R Y) {
      this.X = X;
      this.Y = Y;
    }
    
    public L getX() {
      return X;
    }
    
    public R getY() {
      return Y;
    }
    
    @Override
    public int hashCode() {
      return X.hashCode() ^ Y.hashCode();
    }
    
    @Override
    public boolean equals(Object o) {
      if (!(o instanceof Pair)) return false;
      Pair pairo = (Pair) o;
      return X.equals(pairo.getX()) && Y.equals(pairo.getY());
    }
  }
  
  static final long INFL = 200000000000000000L;
  static final int INF = 2000000000;
  static final boolean DEBUG = true;
  
  static StreamTokenizer in;
  static PrintWriter out;
  
  static void print(String s) {
    print(s, 0);
  }
  
  static void print(String s, int debug) {
    if (debug == 0 || DEBUG) {
      out.print(s);
    }
  }
  
  static void println(String s) {
    println(s, 0);
  }
  
  static void println(String s, int debug) {
    if (debug == 0 || DEBUG) {
      out.println(s);
    }
  }
  
  static void printArray(int[] arr) {
    println(Arrays.toString(arr));
  }
  
  static void printArray(char[] arr) {
    println(Arrays.toString(arr));
  }
  
  static void printArray(String[] arr) {
    println(Arrays.toString(arr));
  }
  
  static void sort(int[] a) {
    Random rnd = new Random();
    for (int i = a.length - 1; i > 0; i--) {
      int index = rnd.nextInt(i);
      a[i] ^= a[index];
      a[index] ^= a[i];
      a[i] ^= a[index];
    }
    Arrays.sort(a);
  }
  
  static char[] inChars;
  static int inCharsInd;
  
  static String next() throws IOException {
    in.nextToken();
    return in.sval;
  }
  
  static char nextChar() throws IOException {
    while (inChars == null || inCharsInd >= inChars.length) {
      inChars = next().toCharArray();
      inCharsInd = 0;
    }
    return inChars[inCharsInd++];
  }
  
  static int nextInt() throws IOException {
    return Integer.valueOf(next());
  }
  
  static double nextDouble() throws IOException {
    return Double.valueOf(next());
  }
  
  static long nextLong() throws IOException {
    return Long.valueOf(next());
  }
  
  private void run() throws IOException {
    solve();
  }
  
  int K, L, M, N, P, globalParent = -1, globalCount = 0;
  int[] visited, parent;

  private void solve() throws IOException {
    N = nextInt();
    parent = new int[N];
    visited = new int[N];
    for (int i = 0; i < N; i++) {
      parent[i] = nextInt()-1;
      if (parent[i] == i) {
        globalParent = parent[i];
      }
    }

    for (int i = 0; i < N; i++) {
      if (visited[i] == 0){
        DFS(i);
      }
    }

    out.println(globalCount);
    for (int i = 0; i < N; i++) {
      out.print((parent[i] + 1) + " ");
    }
  }


  private void DFS(int start){
    visited[start] = 1;
    if (visited[parent[start]] == 1){
      if (globalParent == -1){
        globalParent = start;
      }
      if (parent[start] != globalParent) {
        parent[start] = globalParent;
        globalCount++;
      }
    } else if (visited[parent[start]] == 0){
      DFS(parent[start]);
    }
    visited[start] = 2;
  }
  
}