Berland has managed to repel the flatlanders' attack and is now starting the counter attack.

Flatland has n cities, numbered from 1 to n, and some pairs of them are connected by bidirectional roads. The Flatlandian maps show roads between cities if and only if there is in fact no road between this pair of cities (we do not know whether is it a clever spy-proof strategy or just saving ink). In other words, if two cities are connected by a road on a flatland map, then there is in fact no road between them. The opposite situation is also true: if two cities are not connected by a road on a flatland map, then in fact, there is a road between them.

The berlanders got hold of a flatland map. Now Vasya the Corporal is commissioned by General Touristov to find all such groups of flatland cities, that in each group of cities you can get from any city to any other one, moving along the actual roads. Also the cities from different groups are unreachable from each other, moving along the actual roads. Indeed, destroying such groups one by one is much easier than surrounding all Flatland at once!

Help the corporal complete this task and finally become a sergeant! Don't forget that a flatland map shows a road between cities if and only if there is in fact no road between them. 

Input

The first line contains two space-separated integers n and m (1 ≤ n ≤ 5·105, 0 ≤ m ≤ 106) — the number of cities and the number of roads marked on the flatland map, correspondingly.

Next m lines contain descriptions of the cities on the map. The i-th line contains two integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers of cities that are connected by the i-th road on the flatland map.

It is guaranteed that each pair of cities occurs in the input no more than once.

Output

On the first line print number k — the number of groups of cities in Flatland, such that in each group you can get from any city to any other one by flatland roads. At the same time, the cities from different groups should be unreachable by flatland roads.

On each of the following k lines first print ti (1 ≤ ti ≤ n) — the number of vertexes in the i-th group. Then print space-separated numbers of cities in the i-th group.

The order of printing groups and the order of printing numbers in the groups does not matter. The total sum ti for all k groups must equal n.

Examples

Input

4 4
1 2
1 3
4 2
4 3


Output

2
2 1 4 
2 2 3 


Input

3 1
1 2


Output

1
3 1 2 3 

Note

In the first sample there are roads only between pairs of cities 1-4 and 2-3.

In the second sample there is no road between cities 1 and 2, but still you can get from one city to the other one through city number 3.
import java.io.*;
import java.math.BigInteger;
import java.util.*;



public class Main {

	 static class FastReader 
	    { 
	        final private int BUFFER_SIZE = 1 << 16; 
	        private DataInputStream din; 
	        private byte[] buffer; 
	        private int bufferPointer, bytesRead; 
	  
	        public FastReader() 
	        { 
	            din = new DataInputStream(System.in); 
	            buffer = new byte[BUFFER_SIZE]; 
	            bufferPointer = bytesRead = 0; 
	        } 
	  
	        public FastReader(String file_name) throws IOException 
	        { 
	            din = new DataInputStream(new FileInputStream(file_name)); 
	            buffer = new byte[BUFFER_SIZE]; 
	            bufferPointer = bytesRead = 0; 
	        } 
	  
	        public String readLine() throws IOException 
	        { 
	            byte[] buf = new byte[64]; // line length 
	            int cnt = 0, c; 
	            while ((c = read()) != -1) 
	            { 
	                if (c == '\n') 
	                    break; 
	                buf[cnt++] = (byte) c; 
	            } 
	            return new String(buf, 0, cnt); 
	        } 
	  
	        public int nextInt() throws IOException 
	        { 
	            int ret = 0; 
	            byte c = read(); 
	            while (c <= ' ') 
	                c = read(); 
	            boolean neg = (c == '-'); 
	            if (neg) 
	                c = read(); 
	            do
	            { 
	                ret = ret * 10 + c - '0'; 
	            }  while ((c = read()) >= '0' && c <= '9'); 
	  
	            if (neg) 
	                return -ret; 
	            return ret; 
	        } 
	  
	        public long nextLong() throws IOException 
	        { 
	            long ret = 0; 
	            byte c = read(); 
	            while (c <= ' ') 
	                c = read(); 
	            boolean neg = (c == '-'); 
	            if (neg) 
	                c = read(); 
	            do { 
	                ret = ret * 10 + c - '0'; 
	            } 
	            while ((c = read()) >= '0' && c <= '9'); 
	            if (neg) 
	                return -ret; 
	            return ret; 
	        } 
	  
	        public double nextDouble() throws IOException 
	        { 
	            double ret = 0, div = 1; 
	            byte c = read(); 
	            while (c <= ' ') 
	                c = read(); 
	            boolean neg = (c == '-'); 
	            if (neg) 
	                c = read(); 
	  
	            do { 
	                ret = ret * 10 + c - '0'; 
	            } 
	            while ((c = read()) >= '0' && c <= '9'); 
	  
	            if (c == '.') 
	            { 
	                while ((c = read()) >= '0' && c <= '9') 
	                { 
	                    ret += (c - '0') / (div *= 10); 
	                } 
	            } 
	  
	            if (neg) 
	                return -ret; 
	            return ret; 
	        } 
	  
	        private void fillBuffer() throws IOException 
	        { 
	            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
	            if (bytesRead == -1) 
	                buffer[0] = -1; 
	        } 
	  
	        private byte read() throws IOException 
	        { 
	            if (bufferPointer == bytesRead) 
	                fillBuffer(); 
	            return buffer[bufferPointer++]; 
	        } 
	  
	        public void close() throws IOException 
	        { 
	            if (din == null) 
	                return; 
	            din.close(); 
	        } 
	    } 
	
		public static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	 
		public static long mod = 998244353;
		public static long[] fac;
		public static int N = (int) 1e5;
		public static long[] dp;
		public static List<Integer>[] edges;
		public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub
		FastReader sc = new FastReader();
		
		     int n = sc.nextInt();
		     int m = sc.nextInt();
		     List<Integer>[] edges = new ArrayList[n+1];
		     for(int i=0;i<=n;++i) edges[i] = new ArrayList<>();
		     for(int i=1;i<=m;++i) {
		    	 int u = sc.nextInt();
		    	 int v = sc.nextInt();
		    	 edges[u].add(v);
		    	 edges[v].add(u);
		     }
		     DSU dsu = new DSU(n+1);
		     // to enhance
		     for(int i=1;i<=n;++i) Collections.sort(edges[i]);
		     
		     HashSet<Integer> allComp = new HashSet<>(); //contain one element from each comp
		     
		     for(int i=1;i<=n;++i) {
		    	 Map<Integer,Integer> map = new HashMap<>();
		    	 
		    	 for(int node : edges[i]) {
		    		 /*
		    		  * till now comp. are made from nodes 
		    		  * less than i so there is no use of 
		    		  * node having value greater than i
		    		  */
		    		 if(node >= i) break;
		    		 int root = dsu.findRoot(node);
		    		 /*
		    		  * storing the number of node present in some white comp
		    		  * but connected to i with a black edge
		    		  */
		    		 map.put(root,map.getOrDefault(root,0)+1);
		    	 }
		    	 
		    	 List<Integer> list = new ArrayList<>();
		    	 
		    	 for(int node : allComp) {
		    		 int root = dsu.findRoot(node);
		    		 /*
		    		  * iterating over all comps 
		    		  * if we find any comp having at least one node 
		    		  * which are connected to i by a white edge
		    		  * then we can merge i to that comp
		    		  */
		    		 if(dsu.size[root] > map.getOrDefault(root,0)) {
		    			 dsu.unite(i, root);
		    			 list.add(node);
		    		 }
		    	 }
		    	 /*
		    	  * as now i is merged into diff comp
		    	  * so we delete the elements of that comp
		    	  * and insert i in place of them 
		    	  */
		    	 for(int x : list) allComp.remove(x);
		    	 allComp.add(i);
		     }
		     
		     
		     HashMap<Integer,List<Integer>> map = new HashMap<>();
	         for(int i=1;i<=n;++i) {
	        	 add(map,dsu.findRoot(i),i);
	         }
	         
	         out.println(map.size());
	        for(int x : map.keySet()) {
	        	 out.print(map.get(x).size()+" ");
	        	 for(int v : map.get(x)) {
	        		 out.print(v+" ");
	        	 }
	        	 out.println();
	        }
	        
		out.close();
		}
		
		static void add(Map<Integer,List<Integer>> map, int root, int v) {
			if(map.containsKey(root)) map.get(root).add(v);
			else {
				
				List<Integer> temp = new ArrayList<>();
				temp.add(v);
				map.put(root, temp);
			}
		}
		
		static class DSU {
	        int n;
	        int[] parent, size;

	        public DSU(int v) {
	            n = v;
	            parent = new int[n];
	            size = new int[n];
	            for(int i = 0; i < n; i++) {
	                parent[i] = i;
	                size[i] = 1;
	            }
	        }

	        public int findRoot(int curr) {
	            if(curr == parent[curr]) return curr;
	            return parent[curr] = findRoot(parent[curr]);
	        }

	        public boolean unite(int a, int b) {
	            int rootA = findRoot(a);
	            int rootB = findRoot(b);
	            if(rootA == rootB) return true;
	            if(size[rootA] > size[rootB]) {
	                parent[rootB] = rootA;
	                size[rootA] += size[rootB];
	            }
	            else {
	                parent[rootA] = rootB;
	                size[rootB] += size[rootA];
	            }
	            return false;
	        }

}
}