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:
import java.io.*;
import java.util.*;

public class Solution{
	InputStream is;
	static PrintWriter out;
	String INPUT = "";
	static long mod = (long)1e9+7L;
	
	public void solve(){
		int n = ni(), m = ni();
		ArrayList<Pair>[] al = new ArrayList[n+1];
		boolean[] vis = new boolean[n+1];
		for(int i = 0; i <= n; i++){
			al[i] = new ArrayList<Pair>();
		}

		int y = 1;
		for(int i = 0; i < m; i++){
			int c1 = ni(), c2 = ni();
			vis[c1] = true; vis[c2] = true;

			int x1 = c1, y1 = y;
			int x2 = c2, y2 = y;

			y+=1;

			al[c1].add(new Pair(x1, y1));
			al[c2].add(new Pair(x2, y2)); 
		}

		for(int i = 1; i <= n; i++){
			if(!vis[i]){
				int x1 = i, y1 = y;
				y+=1;
				al[i].add(new Pair(x1, y1));
			}
		}

		for(int i = 1; i <= n; i++){
			out.println(al[i].size());
			for(Pair p : al[i]){
				out.println(p.fs+" "+p.sc);
			}
		}
	}
	

	void run(){
		is = new DataInputStream(System.in);
		out = new PrintWriter(System.out);
		int t=1;while(t-->0)solve();
		out.flush();
	}
	public static void main(String[] args)throws Exception{new Solution().run();}
	long mod(long v, long m){if(v<0){long q=(Math.abs(v)+m-1L)/m;v=v+q*m;}return v%m;}
    long mod(long v){if(v<0){long q=(Math.abs(v)+mod-1L)/mod;v=v+q*mod;}return v%mod;}
	//Fast I/O code is copied from uwi code.
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	private int readByte(){
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	private String ns(){
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	private char[] ns(int n){
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	private char[][] nm(int n, int m){
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	private int[] na(int n){
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	private int ni(){
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	private long nl(){
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	static  int i(long x){return (int)Math.round(x);}
	static class Pair implements Comparable<Pair>{
		long fs,sc;
		Pair(long a,long b){
			fs=a;sc=b;
		}
		public int compareTo(Pair p){
			if(this.fs>p.fs)return 1;
			else if(this.fs<p.fs)return -1;
			else{
				return i(this.sc-p.sc);
			}
			//return i(this.sc-p.sc);
		}
		public String toString(){
			return "("+fs+","+sc+")";
		}	
	}
	
}