Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

4 4
1 2
2 4
1 3
3 4

Sample Output 1

YES

For example, Takahashi's graph has the property mentioned by him if it has the following edges: 1-2, 1-3 and 1-4.
In this case, the number written at every edge will become 2.

Sample Input 2

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

Sample Output 2

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#pragma GCC optimize ("O3")
#include <iostream>
#include <iomanip>
#include <istream>
#include <ostream>
#include <sstream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <list>
#include <stack>
#include <map>
#include <unordered_map>
#include <set>
#include <bitset>
#include <utility>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <string>
#include <ctime>
#include <cctype>
#include <cstdlib>
#define IINF 100000000
#define INF 300000000000000000
#define MOD 1000000007
#define mod 1000000007
#define INT_MAX_ 2147483647
#define EPS (1e-10)
#define REP(i, a, n) for (ll i = a; i < (ll)(n); i++)
#define REPE(i, a, n) for (ll i = a; i <= (ll)(n); i++)
#define rep(i,n)for (ll i = 0; i < (ll)(n); i++)
#define Endl endl
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define mmax(x,y)(x>y?x:y)
#define mmin(x,y)(x<y?x:y)
#define chmax(x,y) x=mmax(x,y)
#define chmin(x,y) x=mmin(x,y)
#define all(x) (x).begin(),(x).end()
#define siz(x) (ll)(x).size()
#define PI acos(-1.0)
#define me memset
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ld;
typedef pair<int,int>Pin;
typedef pair<ll,ll>Pll;
template<class T> using V=vector<T>;
template<typename T> using min_priority_queue = priority_queue<T, vector<T>, greater<T> >;
long long GCD(long long a, long long b) {return b?GCD(b,a%b):a;}
long long LCM(long long a, long long b) {return a/GCD(a,b)*b;}
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
int ddx[8]={-1,0,1,0,1,1,-1,-1};
int ddy[8]={0,-1,0,1,1,-1,1,-1};
ll cmp(pair<ll,Pll> a,pair<ll,Pll> b){
        if(a.fi!=b.fi)
        return a.fi<b.fi;
        else
        return a.se.fi>b.se.fi;
}
//----------------------------------------------------------------------

//----------------------------------------------------------------------
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    //-------------------------------  
    //ll begin_t=clock();
    //freopen("input.txt", "r", stdin);
    //freopen("output.txt", "w", stdout);
    //-------------------------------
    ll n,m;cin>>n>>m;
    V<Pll>ab;
    V<ll>G[100010];
    for(ll i=0;i<m;i++){
        ll a,b;cin>>a>>b;
        if(a>b)swap(a,b);
        ab.pb({a,b});
    }
    sort(all(ab));
    map<Pll,ll>M;
    map<ll,ll>tri;
    for(ll i=0;i<m;i++){
        tri[ab[i].fi]++;
        tri[ab[i].se]++;
    }
    bool abc=1;
    /*for(auto it=M.begin();it!=M.end();it++){
        if(it->second%2!=0){
            abc=0;
            break;
        }
    }*/
    for(auto it =tri.begin();it!=tri.end();it++){
        if(it->second%2!=0){
            abc=0;
            break;
        }
    }
    //if(n%2!=m%2)abc=0;
    if(abc)cout<<"YES"<<endl;
    else cout<<"NO"<<Endl;
    //-------------------------------  
    //fclose(stdin);
    //fclose(stdout);
    //ll end_t=clock();cout<<"time="<<end_t-begin_t<<"ms"<<endl;
    //-------------------------------
    return 0;
}
//----------------------------------------------------------------------