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.IOException;
import java.io.InputStreamReader;
import java.io.Reader;
import java.util.*;

public class D {
    public static void main(String[] args) {
//        Scanner sc = new Scanner(System.in);
        FastScanner sc = new FastScanner();

        int n = sc.nextInt();
        parent = new int[n];
        boolean[] changed = new boolean[n];
        List<Integer>[] adj = new List[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<>();
        }

        for (int child = 0; child < n; child++) {
            parent[child] = sc.nextInt() - 1;
            if (parent[child] != child) {
                adj[child].add(parent[child]);
            }
        }

        SCC scc = new SCC(adj);
        for (List<Integer> component : scc.nodesInScc) {
            if (component.size() > 1) {
                int root = component.get(0);
                parent[root] = root;
                changed[root] = true;
                numChanges++;
            }
        }

        int masterRoot = -1;
        for (int i = 0; i < n; i++) {
            if (parent[i] == i && !changed[i]) {
                masterRoot = i;
                break;
            }
        }
        if (masterRoot == -1) {
            for (int i = 0; i < n; i++) {
                if (parent[i] == i) {
                    masterRoot = i;
                    break;
                }
            }
        }

        for (int i = 0; i < n; i++) {
            if (parent[i] == i && i != masterRoot) {
                parent[i] = masterRoot;
                if (!changed[i]) {
                    numChanges++;
                }
            }
        }

        StringBuilder sb = new StringBuilder();
        sb.append(numChanges).append('\n');
        for (int i = 0; i < n; i++) {
            if (i > 0) sb.append(' ');
            sb.append(parent[i] + 1);
        }
        System.out.println(sb);
    }

    static int n;
    static int[] parent;
    static int numChanges = 0;

    public static class SCC {

        private boolean[] marked;        // marked[v] = has v been visited?
        private int[] id;                // id[v] = id of strong component containing v
        private int[] low;               // low[v] = low number of v
        private int pre;                 // preorder number counter
        private int count;               // number of strongly-connected components
        private Stack<Integer> stack;
        private int[] sccsize;           // size of scc[i], only if desired
        private List<Integer> [] G;      // store underlying graph, only needed for reduce
        List<Integer>[] nodesInScc;

        public SCC(List<Integer> [] G) {
            this.G = G;
            marked = new boolean[G.length];
            stack = new Stack<Integer>();
            id = new int[G.length];
            low = new int[G.length];
            sccsize = new int[G.length];
            nodesInScc = new List[G.length];
            for (int i = 0; i < nodesInScc.length; i++) {
                nodesInScc[i] = new ArrayList<>();
            }
            for (int v = 0; v < G.length; v++) {
                if (!marked[v]) dfs(G, v);
            }
        }

        private void dfs(List<Integer> [] G, int v) {
            marked[v] = true;
            low[v] = pre++;
            int min = low[v];
            stack.push(v);
            for (int w : G[v]) {
                if (!marked[w]) dfs(G, w);
                if (low[w] < min) min = low[w];
            }
            if (min < low[v]) { low[v] = min; return; }
            int w;
            do {
                w = stack.pop();
                id[w] = count;
                sccsize[count]++;
                nodesInScc[count].add(w);
                low[w] = G.length;
            } while (w != v);
            count++;
        }

        // return the number of strongly connected components
        public int count() { return count; }

        // are v and w strongly connected?
        public boolean stronglyConnected(int v, int w) {
            return id[v] == id[w];
        }

        // in which strongly connected component is vertex v?
        public int id(int v) { return id[v]; }

        // how big is the strongly connected component with number #x
        // assert 0 <= x < count
        public int sccsize(int x) { return sccsize[x]; }

        /*
         * Return reduced acyclic graph in which all SCC are collapsed into one node
         */
        @SuppressWarnings("unchecked")
        public List<Integer> [] reduce() {
            Set<Integer> [] set = new Set[count];
            for (int i = 0; i < count; i++)
                set[i] = new HashSet<>();

            for (int i = 0; i < G.length; i++)
                for (Integer j : G[i])
                    if (!stronglyConnected(i, j))
                        set[id[i]].add(id[j]);

            List<Integer>[] r = new List[count];
            for (int i = 0; i < count; i++)
                r[i] = new ArrayList<>(set[i]);

            return r;
        }
    }

    static void shuffle(int[] arr) {
        Random rng = new Random();
        int length = arr.length;
        for (int idx = 0; idx < arr.length; idx++) {
            int toSwap = idx + rng.nextInt(length-idx);
            int tmp = arr[idx];
            arr[idx] = arr[toSwap];
            arr[toSwap] = tmp;
        }
    }

    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        public FastScanner() {
            this(new InputStreamReader(System.in));
        }
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt() {
            return Integer.parseInt(next());
        }
        long nextLong() {
            return Long.parseLong(next());
        }
        double nextDouble() {
            return Double.parseDouble(next());
        }
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
        int[] readIntArray(int n) {
            int[] a = new int[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextInt();
            }
            return a;
        }
        long[] readLongArray(int n) {
            long[] a = new long[n];
            for (int idx = 0; idx < n; idx++) {
                a[idx] = nextLong();
            }
            return a;
        }
    }
}