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.*;

public class Main {

	static long sx = 0, sy = 0, m = (long) (1e9 + 7);

	static ArrayList<Integer>[] a;
	static int[][] dp;
	static long[] farr;
	// static boolean b = true;
	static HashMap<Integer, Integer> hm = new HashMap<>();

	// static TreeMap<Integer, Integer> hm = new TreeMap<>();
	public static PrintWriter out;
	static ArrayList<pair> p = new ArrayList<>();
	static long[] fact = new long[(int) 1e6];
	static boolean b = false;
	static StringBuilder sb = new StringBuilder();
	static boolean cycle = false;
	static long mod = 998244353;
	static int[] col;
	// static HashSet<Integer> p = new HashSet<>();
	static int cnt;

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

		// Scanner scn = new Scanner(System.in);

		Reader scn = new Reader();
		out = new PrintWriter(System.out);

		int n = scn.nextInt();
		int[] a = new int[n + 1];

		for (int i = 1; i <= n; i++)
			a[i] = scn.nextInt();

		int[] c = Arrays.copyOf(a, n + 1);

		dsu d = new dsu(n);

		int cnt = -5; // generic head

		for (int i = 1; i <= n; i++) {

			if (a[i] == i) {

				if (cnt == -5)
					cnt = a[i]; // if a component have head it will have no
								// cycle
				else
					a[i] = -1; // sabhi root ko -1
			}

			else if (!d.merge(i, a[i])) { // this component will have no head so
											// this node at which we got cycle
											// will be the head
				a[i] = -1;
			} // if a component have cycle no head
		}

		int ans = 0;

		for (int i = 1; i <= n; i++) {

			if (a[i] == -1) {

				if (cnt != -5)
					a[i] = cnt;

				else {
					a[i] = i;
					cnt = i;          // if all components had cycle then generic head wont be initialise upar
				}
			}

			if (a[i] != c[i])
				ans++;
		}

		System.out.println(ans);
		for (int i = 1; i <= n; i++)
			System.out.print(a[i] + " ");

	}

	private static class dsu {

		static int[] size;
		static int[] par;

		dsu(int n) {

			size = new int[n + 1];
			par = new int[n + 1];

			for (int i = 1; i <= n; i++) {
				size[i] = 1;
				par[i] = i;
			}
		}

		public static boolean merge(int u, int v) {

			int p1 = find(u);

			int p2 = find(v);

			if (p1 == p2)
				return false;

			if (size[p1] >= size[p2]) {
				size[p1] += size[p2];
				par[p2] = p1;
			}

			else {
				size[p2] += size[p1];
				par[p1] = par[p2];
			}

			return true;
		}

		public static int find(int v) {

			if (v == par[v])
				return v;

			par[v] = find(par[v]);

			return par[v];
		}
	}

	// _________________________TEMPLATE_____________________________________________________________

	// private static int gcd(int a, int b) {
	// if (a == 0)
	// return b;
	//
	// return gcd(b % a, a);
	// }

	// static class comp implements Comparator<Integer> {
	//
	// @Override
	// public int compare(Integer o1, Integer o2) {
	// if (b[o1])
	// return -1;
	// else if (b[o2])
	// return 1;
	// else
	// return (int) (a[o2].size() - a[o1].size());
	// }
	//
	// }

	// public static long pow(long a, long b) {
	//
	// if(b<0)return 0;
	// if (b == 0 || b == 1)
	// return (long) Math.pow(a, b);
	//
	// if (b % 2 == 0) {
	//
	// long ret = pow(a, b / 2);
	// ret = (ret % mod * ret % mod) % mod;
	// return ret;
	// }
	//
	// else {
	// return ((pow(a, b - 1) % mod) * a % mod) % mod;
	// }
	// }

	private static class pair implements Comparable<pair> {

		int tb, tbw;

		pair() {
			tb = tbw = 0;
		}

		@Override
		public int compareTo(pair o) {

			return 1;
		}

	}

	private static String reverse(String s) {

		return new StringBuilder(s).reverse().toString();
	}

	public static class Reader {
		final private int BUFFER_SIZE = 1 << 16;
		private DataInputStream din;
		private byte[] buffer;
		private int bufferPointer, bytesRead;

		public Reader() {
			din = new DataInputStream(System.in);
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public Reader(String file_name) throws IOException {
			din = new DataInputStream(new FileInputStream(file_name));
			buffer = new byte[BUFFER_SIZE];
			bufferPointer = bytesRead = 0;
		}

		public String readLine() throws IOException {
			byte[] buf = new byte[1000000 + 1]; // line length
			int cnt = 0, c;
			while ((c = read()) != -1) {
				if (c == '\n')
					break;
				buf[cnt++] = (byte) c;
			}
			return new String(buf, 0, cnt);
		}

		public int nextInt() throws IOException {
			int ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (neg)
				return -ret;
			return ret;
		}

		public long nextLong() throws IOException {
			long ret = 0;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();
			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');
			if (neg)
				return -ret;
			return ret;
		}

		public double nextDouble() throws IOException {
			double ret = 0, div = 1;
			byte c = read();
			while (c <= ' ')
				c = read();
			boolean neg = (c == '-');
			if (neg)
				c = read();

			do {
				ret = ret * 10 + c - '0';
			} while ((c = read()) >= '0' && c <= '9');

			if (c == '.') {
				while ((c = read()) >= '0' && c <= '9') {
					ret += (c - '0') / (div *= 10);
				}
			}

			if (neg)
				return -ret;
			return ret;
		}

		private void fillBuffer() throws IOException {
			bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
			if (bytesRead == -1)
				buffer[0] = -1;
		}

		private byte read() throws IOException {
			if (bufferPointer == bytesRead)
				fillBuffer();
			return buffer[bufferPointer++];
		}

		public void close() throws IOException {
			if (din == null)
				return;
			din.close();
		}

		public int[] nextIntArray(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nextInt();
			}
			return arr;
		}

		public long[] nextLongArray(int n) throws IOException {
			long[] arr = new long[n];
			for (int i = 0; i < n; i++) {
				arr[i] = nextLong();
			}
			return arr;
		}

		public int[][] nextInt2DArray(int m, int n) throws IOException {
			int[][] arr = new int[m][n];
			for (int i = 0; i < m; i++) {
				for (int j = 0; j < n; j++)
					arr[i][j] = nextInt();
			}
			return arr;
		}
		// kickstart - Solution
		// atcoder - Main
	}

}