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.*;

import java.math.*;

public class P1 {
	static int N;
	static ArrayList<Integer> adj[], adj2[];
	static ArrayList<Long> weight[];
	static ArrayList<Long> preweight[];
	static long depth[];
	static int timeIn[], timeOut[];
	static int time;
	static long a[];

	public static void main(String[] args) {

		new Thread(null, new Runnable() {
			public void run() {
				try {
					solve();
				} catch (Exception e) {
					e.printStackTrace();
				}
			}
		}, "1", 1 << 26).start();

	}

	public static void solve() {

		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

		N = in.nextInt();

		int n = N;

		adj = new ArrayList[N + 1];
		depth = new long[N + 1];

		DisjointSet ds = new DisjointSet(n + 1);

		for (int i = 1; i <= N; i++) {
			adj[i] = new ArrayList<Integer>();
		}
		int root = 0;
		int p[] = new int[N + 1];
		for (int i = 1; i <= N; i++) {

			int a = in.nextInt();
			p[i] = a;
			if (a == i) {
				root = a;
			}
			adj[i].add(a);
			adj[a].add(i);

		}

		int ans = 0;

		for (int i = 1; i <= N; i++) {
			int j = i;
			while (true) {
				if (ds.findSet(j) == ds.findSet(p[j])) {
					break;
				}
				ds.union(j, p[j]);
				j = p[j];
				break;
			}
		}

		for (int i = 1; i <= N; i++) {
			if (p[i] == i) {
				ds.makeparent(i);
			}
		}

		if (root == 0) {
			int tt = ds.findSet(1);
			p[tt] = tt;
			ans++;
			root = tt;
		}

		HashSet<Integer> hs = new HashSet<Integer>();
		int tt = ds.findSet(root);
		hs.add(tt);
		Arrays.fill(depth, -1);

		for (int i = 1; i <= N; i++) {
			int j = ds.findSet(i);
			if (hs.contains(j))
				continue;

			ans++;

			if (p[j] == j) {
				p[j] = root;
			} else {

				an = j;
				an2 = j;

				dfs(j, 0, 0);
				if (p[an2] == an) {
					p[an2] = root;
				} else {
					p[an] = root;
				}
			}
			hs.add(j);
		}
		out.println(ans);
		for (int i = 1; i <= n; i++) {
			out.print(p[i] + " ");
		}

		out.close();

	}

	static int m;
	static long tt;
	static long fac[], ifac[], DR[] = new long[101];
	static int MOD = (int) (1e9 + 7);
	static boolean[] isPrime;
	static int minPrime[];
	static long dp[][];
	static int n;
	static int d, x, y;
	static int an = 0;
	static int an2 = 0;

	static void dfs(int curr, int parent, long pathLength) {
		if (depth[curr] == -1) {
			depth[curr] = pathLength;
			for (int i = 0; i < adj[curr].size(); i++) {
				int child = adj[curr].get(i);
				int edgeWeight = 1;
				if (child == curr) {
					an = curr;
					an2 = curr;
					return;
				}
				if (child != parent && child != curr) {

					if (depth[child] == -1) {

						dfs(child, curr, pathLength + edgeWeight);

					} else {
						an = curr;
						an2 = child;
						return;
					}
				}
			}
		}
	}

	static long lcm(long a, long b) {

		long gg = gcd(a, b);

		double aa = a;

		double cc = (aa / gg) * ((double) b);

		return (long) cc;
	}

	static long gcd(long a, long b) {

		if (b == 0)
			return a;

		return gcd(b, a % b);

	}

	static long mp(long a, long b) {
		long ans = 1;

		while (b != 0) {
			if (b % 2 == 1)
				ans = (ans * a) % MOD;

			a = (a * a) % MOD;
			b /= 2;

		}
		return ans % MOD;
	}

	static void precompute() {
		int i;
		int N = 1000005;
		fac = new long[N];
		ifac = new long[N];
		fac[0] = 1;
		for (i = 1; i < N; i++) {
			fac[i] = (i * fac[i - 1]) % MOD;
		}
		ifac[N - 1] = mp(fac[N - 1], MOD - 2);
		for (i = N - 2; i >= 0; i--) {
			ifac[i] = ((i + 1) * ifac[i + 1]) % MOD;

		}
	}

	public static long combine(int n, int k) {
		if (k > n - k) {
			k = n - k;
		}
		long result = 1;
		for (int i = 0; i < k; i++) {
			result *= (n - i);
			result /= (i + 1);
		}
		return result;
	}

	static long mInv(long A) {

		return mp(A, MOD - 2);

	}

	static long mC(int n, int k) {
		// calculates C(n,k) mod p (assuming p is prime).
		// if(k>n-k)k=n-k;
		long an; // n * (n-1) * ... * (n-k+1)
		if (k <= 0)
			return 1;
		if (n < k)
			return 0;

		an = fac[n] % MOD;
		an *= ifac[n - k];
		an %= MOD;
		an *= ifac[k];
		an %= MOD;

		// numerator / denominator mod p.
		return an;
	}

	static long div(long num) {
		long c = 0;
		for (long i = 1; i <= num; ++i)
			if (num % i == 0)
				c++;
		return c;
	}

	static class DisjointSet {
		int[] rank, parent, size;
		int n;

		DisjointSet(int n) {
			rank = new int[n];
			parent = new int[n];
			size = new int[n + 1];
			this.n = n;
			makeSet();
			size[1] = n;

		}

		int grank(int c) {
			return rank[c];
		}

		void makeSet() {

			for (int i = 0; i < n; i++) { // Initially, all elements are in
											// their
											// own set.
				parent[i] = i;
				rank[i] = 1;
			}
		}

		void makeparent(int a) {

			int p = findSet(a);

			if (parent[a] != a) {
				int temp = parent[p];
				parent[a] = a;
				parent[temp] = a;
			}
			parent[p] = a;
		}

		int findSet(int x) {
			if (parent[x] != x) {

				parent[x] = findSet(parent[x]);
				rank[x] = rank[parent[x]];
			}

			return parent[x];
		}

		int union(int x, int y) {
			int xRoot = findSet(x);
			int yRoot = findSet(y);
			if (xRoot == yRoot) {
				return rank[xRoot];
			}

			parent[xRoot] = yRoot;

			rank[xRoot] += rank[yRoot];
			return 0;
		}
	}

	static class InputReader {

		private InputStream stream;
		private byte[] buf = new byte[8192];
		private int curChar, snumChars;
		private SpaceCharFilter filter;

		public InputReader(InputStream stream) {
			this.stream = stream;
		}

		public int snext() {
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) {
				curChar = 0;
				try {
					snumChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public int nextInt() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = snext();
			}
			int res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = snext();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

		public long nextLong() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			int sgn = 1;
			if (c == '-') {
				sgn = -1;
				c = snext();
			}
			long res = 0;
			do {
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = snext();
			} while (!isSpaceChar(c));
			return res * sgn;
		}

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

		public String readString() {
			int c = snext();
			while (isSpaceChar(c))
				c = snext();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = snext();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public boolean isSpaceChar(int c) {
			if (filter != null)
				return filter.isSpaceChar(c);
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		public interface SpaceCharFilter {
			public boolean isSpaceChar(int ch);
		}
	}
}