Sonya decided to organize an exhibition of flowers. Since the girl likes only roses and lilies, she decided that only these two kinds of flowers should be in this exhibition.

There are n flowers in a row in the exhibition. Sonya can put either a rose or a lily in the i-th position. Thus each of n positions should contain exactly one flower: a rose or a lily.

She knows that exactly m people will visit this exhibition. The i-th visitor will visit all flowers from l_i to r_i inclusive. The girl knows that each segment has its own beauty that is equal to the product of the number of roses and the number of lilies.

Sonya wants her exhibition to be liked by a lot of people. That is why she wants to put the flowers in such way that the sum of beauties of all segments would be maximum possible.

Input

The first line contains two integers n and m (1≤ n, m≤ 10^3) — the number of flowers and visitors respectively.

Each of the next m lines contains two integers l_i and r_i (1≤ l_i≤ r_i≤ n), meaning that i-th visitor will visit all flowers from l_i to r_i inclusive.

Output

Print the string of n characters. The i-th symbol should be «0» if you want to put a rose in the i-th position, otherwise «1» if you want to put a lily.

If there are multiple answers, print any.

Examples

Input

5 3
1 3
2 4
2 5


Output

01100

Input

6 3
5 6
1 4
4 6


Output

110010

Note

In the first example, Sonya can put roses in the first, fourth, and fifth positions, and lilies in the second and third positions;

  * in the segment [1…3], there are one rose and two lilies, so the beauty is equal to 1⋅ 2=2; 
  * in the segment [2…4], there are one rose and two lilies, so the beauty is equal to 1⋅ 2=2; 
  * in the segment [2…5], there are two roses and two lilies, so the beauty is equal to 2⋅ 2=4. 



The total beauty is equal to 2+2+4=8.

In the second example, Sonya can put roses in the third, fourth, and sixth positions, and lilies in the first, second, and fifth positions;

  * in the segment [5…6], there are one rose and one lily, so the beauty is equal to 1⋅ 1=1; 
  * in the segment [1…4], there are two roses and two lilies, so the beauty is equal to 2⋅ 2=4; 
  * in the segment [4…6], there are two roses and one lily, so the beauty is equal to 2⋅ 1=2. 



The total beauty is equal to 1+4+2=7.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.PrintStream;
import java.io.PrintWriter;
public class A {
 
	
	/* 10^(7) = 1s.
	 * ceilVal = (a+b-1) / b */
	
	static final int mod = 1000000007;
	static final long temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	static class Pair implements Comparable<Pair> 
	{
		int first, second;
		public Pair(int first, int second)
		{
			this.first = first;
			this.second = second;
		}
		public int compareTo(Pair o)
		{
			return (int)(o.first - this.first);
		}
	}
	
	static class Tuple implements Comparable<Tuple>{
		long first, second,third;
		public Tuple(long first, long second, long third)
		{
			this.first = first;
			this.second = second;
			this.third = third;
		}
		public int compareTo(Tuple o)
		{
			return (int)(o.third - this.third);
		}
	}
	
	public static class DSU
	{
		int[] parent;
		int[] rank; //Size of the trees is used as the rank
		public DSU(int n)
		{
			parent = new int[n];
			rank = new int[n];
			Arrays.fill(parent, -1);
			Arrays.fill(rank, 1);
		}
		
		public int find(int i) //finding through path compression
		{
			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
		}
		
		public boolean union(int a, int b) //Union Find by Rank 
		{
			a = find(a);
			b = find(b);
			
			if(a == b) return false; //if they are already connected we exit by returning false.
			
			// if a's parent is less than b's parent
			if(rank[a] < rank[b])
			{
				//then move a under b
				parent[a] = b;
			}
			//else if rank of j's parent is less than i's parent
			else if(rank[a] > rank[b])
			{
				//then move b under a
				parent[b] = a;
			}
			//if both have the same rank.
			else
			{
				//move a under b (it doesnt matter if its the other way around.
				parent[b] = a;
				rank[a] = 1 + rank[a];
			}
			return true;
		}	
	}

	static class Reader {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
	public static boolean isPrime(long n) 
	{
		if(n == 1)
		{
			return false;
		}
		for(int i = 2;i*i<=n;i++)
		{
			if(n%i == 0)
			{
				return false;
			}
		}
		return true;
}
	
	public static List<Integer> SieveList(int n)
	{
		boolean prime[] = new boolean[n+1];
		Arrays.fill(prime, true);
		
		ArrayList<Integer> l = new ArrayList<>();
		for (int p=2; p*p<=n; p++) 
		{ 
			if (prime[p] == true) 
			{ 	
			    for(int i=p*p; i<=n; i += p) 
				{
				    prime[i] = false; 
				}	
			} 
		} 
 
		for (int p=2; p<=n; p++) 
		{
		    if (prime[p] == true)
		    {
		       l.add(p); 
		    }
		}
		
		return l;
	}
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
		public static int lcm(int a, int b)
	    {
	        return (a / gcd(a, b)) * b;
	    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
		public static long LongLCM(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
		//Count the number of coprime's upto N
		public static long phi(long n)  //euler totient/phi function
	    { 
	       long ans = n;
//	       for(long i = 2;i*i<=n;i++)
//	       {
//	    	   if(n%i == 0)
//	    	   {
//	    		   while(n%i == 0) n/=i;
//	    		   ans -= (ans/i);
//	    	   }
//	       }
//	       
//	       if(n > 1)
//	       {
//	    	   ans -= (ans/n);
//	       }
	       
	       for(long i = 2;i<=n;i++)
	       {
	    	   if(isPrime(i))
	    	   {
	    		   ans -= (ans/i);
	    	   }
	       }
	       
	       return ans;
	       
	    }	
	
	public static long fastPow(long x, long n) 
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	   public static long modPow(long x, long y, long p)
	    {
	        long res = 1;
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCrModP(long n, long r,long p)
	    {
	    	if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long fact(long n) {
	    	  long[] fac = new long[(int)(n) + 1];
		        fac[0] = 1;
		 
		        for (long i = 1; i <= n; i++)
		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
		        
		        return fac[(int)(n)];
		}
	    
	    public static long nCr(long n, long k) 
	    {
			long ans = 1;
			for(long i = 0;i<k;i++)
			{
				ans *= (n-i);
				ans /= (i+1);
			}
			return ans;
		}
 
		public static void Sort(long[] a) 
		{
			List<Long> l = new ArrayList<>();
			for (long i : a) l.add(i);
			Collections.sort(l);
//			Collections.reverse(l);  //Use to Sort decreasingly 
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		public static void ssort(char[] a)
		{
			List<Character> l = new ArrayList<>();
			for (char i : a) l.add(i);
			Collections.sort(l);
			for (int i=0; i<a.length; i++) a[i]=l.get(i);
		}
		
		//Modular Operations for Addition and Multiplication. 
		   public static long perfomMod(long x)
		   {
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		  
		   public static long subMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)-perfomMod(b));
		   }
		   
		   public static long mulMod(long  a, long b)
		   {
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		   
		   /*
		    * 
		    * >=         <=
		   0 1 2 3 4 5 6 7  
		   5 5 5 6 6 6 7 7
		   lower_bound for 6 at index 3 (>=)
		   upper_bound for 6 at index 6(To get six reduce by one) (<=)
		   */
		
		   public static int LowerBound(int a[], int x) 
		   {
				  int l=-1,r=a.length;
				  while(l+1<r) {
				    int m=(l+r)>>>1;
				    if(a[m]>=x) r=m;
				    else l=m;
				  }
				  return r;
			}
		
		   
			public static int UpperBound(int a[], int x) 
			{
			    int l=-1, r=a.length;
			    while(l+1<r) {
			       int m=(l+r)>>>1;
			       if(a[m]<=x) l=m;
			       else r=m;
			    }
			    return l+1;
			}
	
		
	public static void main(String[] args) throws Exception 
	{
		Reader sc = new Reader();
		PrintWriter fout = new PrintWriter(System.out);

		int n = sc.nextInt(), m = sc.nextInt();
		int[] l = new int[m];
		int[] r = new int[m];
		for(int i = 0;i<m;i++)
		{
			l[i] = sc.nextInt();
			r[i] = sc.nextInt();
		}
		
		StringBuilder app = new StringBuilder();
		for(int i = 0;i<n;i++)
		{
			if(i%2 == 0)
			{
				app.append("0");
			}
			else
			{
				app.append("1");
			}
		}
		
		fout.println(app.toString());
		
		fout.close();
	}
}
