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_Round363Div2_D {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		Printer pr = new Printer(System.out);

		int n = sc.nextInt();

		UnionFind uf = new UnionFind(n);

		int[] p = new int[n];
		List<Integer> roots = new ArrayList<>();
		for (int i = 0; i < n; i++) {
			p[i] = sc.nextInt() - 1;
			if (i == p[i]) {
				roots.add(i);
			}
			uf.unite(i, p[i]);
		}

		int ret = 0;
		if (roots.size() == 0) {
			Queue<Integer> q = new ArrayDeque<>();
			q.add(0);
			boolean[] used = new boolean[n];
			used[0] = true;
			while (!q.isEmpty()) {
				int e = q.remove();

				int ne = p[e];
				if (used[ne]) {
					ret++;
					roots.add(ne);
					p[ne] = ne;
					break;
				} else {
					q.add(ne);
					used[ne] = true;
				}
			}
		}

		while (roots.size() > 1) {
			int tmp = roots.get(roots.size() - 1);
			p[tmp] = roots.get(0);
			uf.unite(tmp, roots.get(0));
			ret++;
			roots.remove(roots.size() - 1);
		}

		if (uf.count() > 1) {
			int root = roots.get(0);
			boolean[] used = new boolean[n];
			for (int i = 0; i < n; i++) {
				if (uf.same(root, i)) {
					continue;
				}

				Queue<Integer> q = new ArrayDeque<>();
				q.add(i);
				used[i] = true;
				while (!q.isEmpty()) {
					int e = q.remove();

					int ne = p[e];
					if (used[ne]) {
						ret++;
						p[ne] = root;
						uf.unite(root, ne);
						break;
					} else {
						q.add(ne);
						used[ne] = true;
					}
				}
			}
		}

		pr.println(ret);
		for (int i = 0; i < n; i++) {
			if (i > 0) {
				pr.print(" ");
			}
			pr.print(p[i] + 1);
		}
		pr.println();

		pr.close();
		sc.close();
	}

	@SuppressWarnings("unused")
	private static class UnionFind {
		int n;
		// cnt : 異なる集合の数
		int cnt;
		// parent[x] : 0～n-1 の場合、要素xのroot要素
		//           : -1～-n の場合、自分自身がroot要素、
		//                            -parent[x]でxを含む集合の要素数
		int[] parent;

		UnionFind(int n) {
			this.n = n;
			cnt = n;
			parent = new int[n];
			Arrays.fill(parent, -1);
		}

		// xのrootを求める
		int find(int x) {
			if (parent[x] < 0) {
				return x;
			} else {
				return parent[x] = find(parent[x]);
			}
		}

		// xとyが同じ集合に属するのか
		boolean same(int x, int y) {
			return find(x) == find(y);
		}

		// xとyの属する集合を併合する
		void unite(int x, int y) {
			x = find(x);
			y = find(y);
			if (x == y) {
				return;
			}

			cnt--;
			// 要素数が大きい集合をrootにする(Quick Find Weighted?)
			if (parent[x] > parent[y]) {
				parent[y] += parent[x];
				parent[x] = y;
			} else {
				parent[x] += parent[y];
				parent[y] = x;
			}

			return;
		}

		// 要素xを含む集合の要素数
		int count(int x) {
			return -parent[find(x)];
		}

		// 異なる集合の数
		int count() {
			return cnt;
		}
	}

	@SuppressWarnings("unused")
	private static class Scanner {
		BufferedReader br;
		Iterator<String> it;

		Scanner (InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}

		String next() throws RuntimeException  {
			try {
				if (it == null || !it.hasNext()) {
//					it = Arrays.asList(br.readLine().split(" ")).iterator();
					it = Arrays.asList(br.readLine().split("\\p{javaWhitespace}+")).iterator();
				}
				return it.next();
			} catch (IOException e) {
				throw new IllegalStateException();
			}
		}

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

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

		float nextFloat() throws RuntimeException {
			return Float.parseFloat(next());
		}

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

		void close() {
			try {
				br.close();
			} catch (IOException e) {
//				throw new IllegalStateException();
			}
		}
	}

	private static class Printer extends PrintWriter {
		Printer(PrintStream out) {
			super(out);
		}
	}
}