Problem: Ivan is a novice painter. He has n dyes of different colors. He also knows exactly m pairs of colors which harmonize with each other.

Ivan also enjoy playing chess. He has 5000 rooks. He wants to take k rooks, paint each of them in one of n colors and then place this k rooks on a chessboard of size 10^{9} × 10^{9}.

Let's call the set of rooks on the board connected if from any rook we can get to any other rook in this set moving only through cells with rooks from this set. Assume that rooks can jump over other rooks, in other words a rook can go to any cell which shares vertical and to any cell which shares horizontal.

Ivan wants his arrangement of rooks to have following properties:

  * For any color there is a rook of this color on a board;
  * For any color the set of rooks of this color is connected;
  * For any two different colors a b union of set of rooks of color a and set of rooks of color b is connected if and only if this two colors harmonize with each other.



Please help Ivan find such an arrangement.

Input

The first line of input contains 2 integers n, m (1 ≤ n ≤ 100, 0 ≤ m ≤ min(1000,    (n(n-1))/(2))) — number of colors and number of pairs of colors which harmonize with each other.

In next m lines pairs of colors which harmonize with each other are listed. Colors are numbered from 1 to n. It is guaranteed that no pair occurs twice in this list.

Output

Print n blocks, i-th of them describes rooks of i-th color.

In the first line of block print one number a_{i} (1 ≤ a_{i} ≤ 5000) — number of rooks of color i. In each of next a_{i} lines print two integers x and y (1 ≤ x,    y ≤ 10^{9}) — coordinates of the next rook.

All rooks must be on different cells.

Total number of rooks must not exceed 5000.

It is guaranteed that the solution exists.

Examples

Input

3 2
1 2
2 3


Output

2
3 4
1 4
4
1 2
2 2
2 4
5 4
1
5 1


Input

3 3
1 2
2 3
3 1


Output

1
1 1
1
1 2
1
1 3


Input

3 1
1 3


Output

1
1 1
1
2 2
1
3 1

Note

Rooks arrangements for all three examples (red is color 1, green is color 2 and blue is color 3).

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Solution (in JAVA):

Solution:

/**
 * Date: 24 Oct, 2018
 * Link:
 *
 * @author Prasad-Chaudhari
 * @linkedIn: https://www.linkedin.com/in/prasad-chaudhari-841655a6/
 * @git: https://github.com/Prasad-Chaudhari
 */
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.LinkedList;

public class newProgram6 {

    public static void main(String[] args) throws IOException {
        // TODO code application logic here
        FastIO in = new FastIO();
        int n = in.ni();
        int m = in.ni();
        int xs[] = new int[m];
        int ys[] = new int[m];
        for (int i = 0; i < m; i++) {
            xs[i] = in.ni();
            ys[i] = in.ni();
        }
        int count[] = new int[n + 1];
        boolean bb[][] = new boolean[n + 1][5001];
        Point[][] p = new Point[n + 1][5000];
        for (int i = 0; i < m; i++) {
            int x = xs[i];
            int y = ys[i];
            p[x][i] = new Point(x, i + 1);
            p[y][i] = new Point(y, i + 1);
            count[x]++;
            count[y]++;
        }
        int end = (int) 1e9;
        for (int i = 1; i <= n; i++) {
            if (count[i] == 0) {
                System.out.println("1");
                System.out.println(i + " " + end--);
            } else {
                System.out.println(count[i]);
                for (int j = 0; j < 5000; j++) {
                    if (p[i][j] != null) {
                        System.out.println(p[i][j].x + " " + p[i][j].y);
                    }
                }
            }
        }
    }

    static class Point {

        int x;
        int y;

        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }

    }

    static class UnionFind {

        private int noOfComponents, n;
        private int component[], size[];
        private ArrayList<LinkedList<Integer>> members;

        public UnionFind(int p) {
            n = p;
            noOfComponents = n;
            component = new int[n + 1];
            size = new int[n + 1];
            members = new ArrayList<>();
            for (int i = 0; i <= n; i++) {
                component[i] = i;
                size[i] = 1;
                members.add(new LinkedList<Integer>());
                members.get(i).add(i);
            }
        }

        public int find(int k) {
            return component[k];
        }

        public void union(int a, int b) {
            if (size[a] > size[b]) {
                a = a ^ b;
                b = a ^ b;
                a = a ^ b;
            }
            LinkedList<Integer> membersofa = members.get(a);
            LinkedList<Integer> membersofb = members.get(b);
            for (int i = 0; i < size[a]; i++) {
                int member = membersofa.removeFirst();
                component[member] = component[b];
                membersofb.add(member);
            }
            size[b] = size[b] + size[a];
            noOfComponents--;
        }

        public int noOfComponents() {
            return noOfComponents;
        }
    }

    static class FastIO {

        private final BufferedReader br;
        private final BufferedWriter bw;
        private String s[];
        private int index;
        private StringBuilder sb;

        public FastIO() throws IOException {
            br = new BufferedReader(new InputStreamReader(System.in));
            bw = new BufferedWriter(new OutputStreamWriter(System.out, "UTF-8"));
            s = br.readLine().split(" ");
            sb = new StringBuilder();
            index = 0;
        }

        public int ni() throws IOException {
            return Integer.parseInt(nextToken());
        }

        public double nd() throws IOException {
            return Double.parseDouble(nextToken());
        }

        public long nl() throws IOException {
            return Long.parseLong(nextToken());
        }

        public String next() throws IOException {
            return nextToken();
        }

        public String nli() throws IOException {
            try {
                return br.readLine();
            } catch (IOException ex) {

            }
            return null;
        }

        public int[] gia(int n) throws IOException {
            int a[] = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public int[] gia(int n, int start, int end) throws IOException {
            validate(n, start, end);
            int a[] = new int[n];
            for (int i = start; i < end; i++) {
                a[i] = ni();
            }
            return a;
        }

        public double[] gda(int n) throws IOException {
            double a[] = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = nd();
            }
            return a;
        }

        public double[] gda(int n, int start, int end) throws IOException {
            validate(n, start, end);
            double a[] = new double[n];
            for (int i = start; i < end; i++) {
                a[i] = nd();
            }
            return a;
        }

        public long[] gla(int n) throws IOException {
            long a[] = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }

        public long[] gla(int n, int start, int end) throws IOException {
            validate(n, start, end);
            long a[] = new long[n];
            for (int i = start; i < end; i++) {
                a[i] = nl();
            }
            return a;
        }

        public int[][] gg(int n, int m) throws IOException {
            int adja[][] = new int[n + 1][];
            int from[] = new int[m];
            int to[] = new int[m];
            int count[] = new int[n + 1];
            for (int i = 0; i < m; i++) {
                from[i] = ni();
                to[i] = ni();
                count[from[i]]++;
                count[to[i]]++;
            }
            for (int i = 0; i <= n; i++) {
                adja[i] = new int[count[i]];
            }
            for (int i = 0; i < m; i++) {
                adja[from[i]][--count[from[i]]] = to[i];
                adja[to[i]][--count[to[i]]] = from[i];
            }
            return adja;
        }

        public void print(String s) throws IOException {
            bw.write(s);
            bw.flush();
        }

        public void println(String s) throws IOException {
            bw.write(s);
            bw.newLine();
            bw.flush();
        }

        private String nextToken() throws IndexOutOfBoundsException, IOException {
            if (index == s.length) {
                s = br.readLine().split(" ");
                index = 0;
            }
            return s[index++];
        }

        private void validate(int n, int start, int end) {
            if (start < 0 || end >= n) {
                throw new IllegalArgumentException();
            }
        }
    }

    static class Data implements Comparable<Data> {

        int a, b;

        public Data(int a, int b) {
            this.a = a;
            this.b = b;
        }

        @Override
        public int compareTo(Data o) {
            if (a == o.a) {
                return Integer.compare(b, o.b);
            }
            return Integer.compare(a, o.a);
        }
    }
}