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.BufferedReader;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class D363_FixTree {
	static int parent[];
	
	public static void main(String args[]) throws FileNotFoundException {
		InputReader in = new InputReader(System.in);
		OutputStream outputStream = System.out;
		PrintWriter out = new PrintWriter(outputStream);
		StringBuilder sb = new StringBuilder();
		// ----------My Code----------
		/*
		 * int n=in.nextInt(); int n1,n2; int a[],b[]; n1=in.nextInt();
		 * a=in.nextIntArray(n1, 0); n2=in.nextInt(); b=in.nextIntArray(n2, 0);
		 */

        int n = in.nextInt();
        int[] p = new int[n];
        parent = new int[n];
        int root = -1;
        for (int i = 0; i < n; i++) parent[i] = i;
        for (int i = 0; i < n; i++) {
            p[i] = in.nextInt() - 1;
            if(p[i] == i)
                root = i;
        }
        int count = 0;
        for (int i = 0; i < n; i++) {
            if(find(i) != find(p[i])) {
                union(p[i], i);
            } else {
                if(root == -1)
                    root = find(i); 
                if(p[i] != root) {
                    p[i] = root;
                    count++;
                }
            }
        }
        sb.append(count).append("\n");
       
        for (int i = 0; i < n; i++)
            sb.append((p[i]+1) + " ");
        out.println(sb);
     

		// ---------------The End------------------

        out.close();
    }
    
    public static void union(int x, int y) {
        int px = find(x);
        int py = find(y);
        parent[py] = px;
    }
    
    public static int find(int x) {
        if(x == parent[x])
            return x;
        return (parent[x] = find(parent[x]));
    }

	// ---------------Extra Methods------------------

	public static long pow(long x, long n, long mod) {
		long res = 1;
		x %= mod;
		while (n > 0) {
			if (n % 2 == 1) {
				res = (res * x) % mod;
			}
			x = (x * x) % mod;
			n /= 2;
		}
		return res;
	}

	public static boolean isPal(String s) {
		for (int i = 0, j = s.length() - 1; i <= j; i++, j--) {
			if (s.charAt(i) != s.charAt(j))
				return false;
		}
		return true;
	}

	public static String rev(String s) {
		StringBuilder sb = new StringBuilder(s);
		sb.reverse();
		return sb.toString();
	}

	public static long gcd(long x, long y) {
		if (x % y == 0)
			return y;
		else
			return gcd(y, x % y);
	}

	public static int gcd(int x, int y) {
		if (x % y == 0)
			return y;
		else
			return gcd(y, x % y);
	}

	public static long gcdExtended(long a, long b, long[] x) {

		if (a == 0) {
			x[0] = 0;
			x[1] = 1;
			return b;
		}
		long[] y = new long[2];
		long gcd = gcdExtended(b % a, a, y);

		x[0] = y[1] - (b / a) * y[0];
		x[1] = y[0];

		return gcd;
	}

	public static int abs(int a, int b) {
		return (int) Math.abs(a - b);
	}

	public static long abs(long a, long b) {
		return (long) Math.abs(a - b);
	}

	public static int max(int a, int b) {
		if (a > b)
			return a;
		else
			return b;
	}

	public static int min(int a, int b) {
		if (a > b)
			return b;
		else
			return a;
	}

	public static long max(long a, long b) {
		if (a > b)
			return a;
		else
			return b;
	}

	public static long min(long a, long b) {
		if (a > b)
			return b;
		else
			return a;
	}
	// ---------------Extra Methods------------------

	public static void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	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 char nextChar() {
			return next().charAt(0);
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public int[] nextIntArray(int n, int f) {
			if (f == 0) {
				int[] arr = new int[n];
				for (int i = 0; i < n; i++) {
					arr[i] = nextInt();
				}
				return arr;
			} else {
				int[] arr = new int[n + 1];
				for (int i = 1; i < n + 1; i++) {
					arr[i] = nextInt();
				}
				return arr;
			}
		}

		public long[] nextLongArray(int n, int f) {
			if (f == 0) {
				long[] arr = new long[n];
				for (int i = 0; i < n; i++) {
					arr[i] = nextLong();
				}
				return arr;
			} else {
				long[] arr = new long[n + 1];
				for (int i = 1; i < n + 1; i++) {
					arr[i] = nextLong();
				}
				return arr;
			}

		}

		public double[] nextDoubleArray(int n, int f) {
			if (f == 0) {
				double[] arr = new double[n];
				for (int i = 0; i < n; i++) {
					arr[i] = nextDouble();
				}
				return arr;
			} else {
				double[] arr = new double[n + 1];
				for (int i = 1; i < n + 1; i++) {
					arr[i] = nextDouble();
				}
				return arr;
			}
		}

	}

	static class Pair implements Comparable<Pair> {
		int a, b;

		Pair(int a, int b) {
			this.a = a;
			this.b = b;
		}

		public int compareTo(Pair o) {
			// TODO Auto-generated method stub
			if (this.a != o.a)
				return Integer.compare(this.a, o.a);
			else
				return Integer.compare(this.b, o.b);
		}

		public boolean equals(Object o) {
			if (o instanceof Pair) {
				Pair p = (Pair) o;
				return p.a == a && p.b == b;
			}
			return false;
		}

		public int hashCode() {
			return new Integer(a).hashCode() * 31 + new Integer(b).hashCode();
		}

	}

}