Problem: Takahashi is not good at problems about trees in programming contests, and Aoki is helping him practice.

First, Takahashi created a tree with N vertices numbered 1 through N, and wrote 0 at each edge.

Then, Aoki gave him M queries. The i-th of them is as follows:

* Increment the number written at each edge along the path connecting vertices a_i and b_i, by one.



After Takahashi executed all of the queries, he told Aoki that, for every edge, the written number became an even number. However, Aoki forgot to confirm that the graph Takahashi created was actually a tree, and it is possible that Takahashi made a mistake in creating a tree or executing queries.

Determine whether there exists a tree that has the property mentioned by Takahashi.

Constraints

* 2 ≤ N ≤ 10^5
* 1 ≤ M ≤ 10^5
* 1 ≤ a_i,b_i ≤ N
* a_i ≠ b_i

Input

Input is given from Standard Input in the following format:


N M
a_1 b_1
:
a_M b_M


Output

Print `YES` if there exists a tree that has the property mentioned by Takahashi; print `NO` otherwise.

Examples

Input

4 4
1 2
2 4
1 3
3 4


Output

YES


Input

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


Output

NO

Solution (in JAVA):

Solution:
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.*;
import java.math.BigInteger;
import static java.util.Comparator.*;

public class Main {	
	
	public static void main(String[] args) throws Exception {
		
		//int N = scan.nextInt();
		//long L = scan.nextLong();
		//double D = scan.nextDouble();
		//char[] A = scan.next().toCharArray();
		
		
		int N = scan.nextInt();
		int M = scan.nextInt();
		int[] c = new int[N];
		for(int i=0; i<M; i++){
			int a = scan.nextInt()-1;
			int b = scan.nextInt()-1;
			c[a]++;
			c[b]++;
		}
		boolean exist=true;
		for(int i: c)if(i%2==1)exist=false;
	
		out.println(exist? "YES" : "NO");
		out.flush();
	}

	static FastScanner scan = new FastScanner();
	static PrintWriter out = new PrintWriter(System.out);			 
	static long MOD = 1_000_000_007;

	static int[] dx = {1,-1,0,0};
	static int[] dy = {0,0,1,-1};
	static int[] dx8 = {1,-1,0,0,1,1,-1,-1};
	static int[] dy8 = {0,0,1,-1,1,-1,1,-1};

	//GCD最大公約数を返す
	static long gcd(long a, long b){
		if(b == 0) return a;
		return gcd(b, a%b);
	}

	//aとbの最小公倍数を返す
	static long lcm (long a, long b) {
		return ((a/gcd(a,b))*b);
	}
	
	//入力
	//https://qiita.com/p_shiki37/items/a0f6aac33bf60f5f65e4
	static class FastScanner {
	    private final InputStream in = System.in;
	    private final byte[] buffer = new byte[1024];
	    private int ptr = 0;
	    private int buflen = 0;
	    private boolean hasNextByte() {
	        if (ptr < buflen) {
	            return true;
	        }else{
	            ptr = 0;
	            try {
	                buflen = in.read(buffer);
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	            if (buflen <= 0) {
	                return false;
	            }
	        }
	        return true;
	    }
	    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
	    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
	    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
	    public String next() {
	        if (!hasNext()) throw new NoSuchElementException();
	        StringBuilder sb = new StringBuilder();
	        int b = readByte();
	        while(isPrintableChar(b)) {
	            sb.appendCodePoint(b);
	            b = readByte();
	        }
	        return sb.toString();
	    }
	    public long nextLong() {
	        if (!hasNext()) throw new NoSuchElementException();
	        long n = 0;
	        boolean minus = false;
	        int b = readByte();
	        if (b == '-') {
	            minus = true;
	            b = readByte();
	        }
	        if (b < '0' || '9' < b) {
	            throw new NumberFormatException();
	        }
	        while(true){
	            if ('0' <= b && b <= '9') {
	                n *= 10;
	                n += b - '0';
	            }else if(b == -1 || !isPrintableChar(b)){
	                return minus ? -n : n;
	            }else{
	                throw new NumberFormatException();
	            }
	            b = readByte();
	        }
	    }
	    public int nextInt() {
	        long nl = nextLong();
	        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
	        return (int) nl;
	    }
	    public double nextDouble() { return Double.parseDouble(next());}
	}
}