Problem: Berland Government decided to improve relations with neighboring countries. First of all, it was decided to build new roads so that from each city of Berland and neighboring countries it became possible to reach all the others. There are n cities in Berland and neighboring countries in total and exactly n - 1 two-way roads. Because of the recent financial crisis, the Berland Government is strongly pressed for money, so to build a new road it has to close some of the existing ones. Every day it is possible to close one existing road and immediately build a new one. Your task is to determine how many days would be needed to rebuild roads so that from each city it became possible to reach all the others, and to draw a plan of closure of old roads and building of new ones.

Input

The first line contains integer n (2 ≤ n ≤ 1000) — amount of cities in Berland and neighboring countries. Next n - 1 lines contain the description of roads. Each road is described by two space-separated integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — pair of cities, which the road connects. It can't be more than one road between a pair of cities. No road connects the city with itself.

Output

Output the answer, number t — what is the least amount of days needed to rebuild roads so that from each city it became possible to reach all the others. Then output t lines — the plan of closure of old roads and building of new ones. Each line should describe one day in the format i j u v — it means that road between cities i and j became closed and a new road between cities u and v is built. Cities are numbered from 1. If the answer is not unique, output any.

Examples

Input

2
1 2


Output

0


Input

7
1 2
2 3
3 1
4 5
5 6
6 7


Output

1
3 1 3 7

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Map;
import java.util.StringTokenizer;

public class D {
        private Map<Long, Node> map = new HashMap<>();
        static int dsu[];
    class Node {
        long data;
        Node parent;
        int rank;
    }

    /**
     * Create a set with only one element.
     */
    public void makeSet(long data) {
        Node node = new Node();
        node.data = data;
        node.parent = node;
        node.rank = 0;
        map.put(data, node);
    }

    /**
     * Combines two sets together to one.
     * Does union by rank
     *
     * @return true if data1 and data2 are in different set before union else false.
     */
    public boolean union(long data1, long data2) {
        Node node1 = map.get(data1);
        Node node2 = map.get(data2);

        Node parent1 = findSet(node1);
        Node parent2 = findSet(node2);

        //if they are part of same set do nothing
        if (parent1.data == parent2.data) {
            return false;
        }

        //else whoever's rank is higher becomes parent of other
        if (parent1.rank >= parent2.rank) {
            //increment rank only if both sets have same rank
            parent1.rank = (parent1.rank == parent2.rank) ? parent1.rank + 1 : parent1.rank;
            parent2.parent = parent1;
        } else {
            parent1.parent = parent2;
        }
        return true;
    }

    /**
     * Finds the representative of this set
     */
    public long findSet(long data) {
        return findSet(map.get(data)).data;
    }

    /**
     * Find the representative recursively and does path
     * compression as well.
     */
    private Node findSet(Node node) {
        Node parent = node.parent;
        if (parent == node) {
            return parent;
        }
        node.parent = findSet(node.parent);
        return node.parent;
    }

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.flush();
    }
    static class TaskB{
        void solve(int test , InputReader in , PrintWriter out) {
            
            int n = in.nextInt();
            D ds = new D();
            ArrayList<pair> rem = new ArrayList();
            for (int i = 1; i <= n; i++) 
                ds.makeSet(i);
            
            for (int i = 0 , a , b; i < n-1; i++) {
                a = in.nextInt();
                b = in.nextInt();
                if(ds.findSet(a) == ds.findSet(b))
                    rem.add(new pair(a,b));
                else {
//                    ds.findSet(a = ds.findSet(b);
                    ds.union(a, b);
                }
            }
            ArrayList<Integer> c = new ArrayList();
            for (int i = 1; i <= n; i++) {
                if(ds.findSet(i) == i)
                    c.add(i);
            }
            out.println(c.size()-1);
            for (int i = 0; i < c.size()-1; i++) {
                out.printf("%d %d %d %d \n",rem.get(i).first , rem.get(i).second,c.get(i),c.get(i+1));
            }
            }
        }
        public static class pair{
        int first , second;
        pair(int first,int second){
            this.first = first;
            this.second = second;
        }
    }
       static class InputReader{
       public BufferedReader reader;
       public StringTokenizer tk;
       public InputReader(InputStream stream){
           reader = new BufferedReader(new InputStreamReader(stream));
           tk = null;
       }
       String next(){
           while (tk == null || !tk.hasMoreTokens()) {
               try{
                   tk = new StringTokenizer(reader.readLine());
               }
               catch(IOException e){
                   throw new RuntimeException();
               }
           }
           return tk.nextToken();
       }
       int nextInt(){
           return Integer.parseInt(next());
       }
       long nextLong(){
           return Long.parseLong(next());
       }        
   } 
}