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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*
import java.io.PrintWriter

val pw = PrintWriter(System.out)
val MOD = 1000000007L
val INF = 2147483647
val LINF = 9223372036854775807L

fun main(args: Array<String>){
    solve()
    pw.flush()    
}

fun solve(){
    val (n, m) = nextIntList()
    val count = IntArray(n) { 0 }
    var ans = "YES"
    
    repeat(m) { 
        val (a, b) = nextIntList()
        count[a - 1] += 1
        count[b - 1] += 1
    }
    for (i in 0 until n) {
        if(count[i] % 2 != 0){
            ans = "NO"
            break;
        }
    }

    println(ans)

}


// Print
fun println(v: String){
    pw.println(v)
}
fun print(v: String){
    pw.print(v)
}

// Read
fun next() = readLine()!!
fun nextInt() = next().toInt()
fun nextLong() = next().toLong()
fun nextDouble() = next().toDouble()
fun nextList() = next().split(" ")
fun nextIntList() = next().split(" ").map{ it.toInt() }
fun nextLongList() = next().split(" ").map{ it.toLong() }
fun nextDoubleList() = next().split(" ").map{ it.toDouble() }
fun nextAryln(n: Int) = Array(n){ next() }
fun nextIntAryln(n: Int) = IntArray(n){ nextInt() }
fun nextLongAryln(n: Int) = LongArray(n){ nextLong() }
fun nextDoubleAryln(n: Int) = DoubleArray(n) { nextDouble() }

// Math
fun abs(n: Long) : Long = Math.abs(n)
fun abs(n: Double) : Double = Math.abs(n)
fun max(a: Long = -LINF, b: Long = -LINF, c: Long = -LINF, d: Long = -LINF, e: Long = -LINF): Long = listOf(a, b, c, d, e).max()!!.toLong()
fun min(a: Long = LINF, b: Long = LINF, c: Long = LINF, d: Long = LINF, e: Long = LINF): Long = listOf(a, b, c, d, e).min()!!.toLong()
fun prime(from: Long, to: Long = from) : List<Long>{
    return (from..to).filter{ i ->
        val max = Math.sqrt(i.toDouble()).toLong()
        (2..max).all{ j -> i % j != 0L}
    }
}
fun gcd(a: Long, b: Long) : Long = if(a % b == 0L) b else gcd(b, (a % b))
fun lcm(a: Long, b: Long) : Long = a / gcd(a, b) * b
fun modpow(a: Long, n: Long, p:Long = MOD) : Long {
    var res = 1L
    var ar = a
    var nr = n
    while(nr > 0){
        if((nr and 1) == 1L) res = res * ar % p
        ar = ar * ar % p
        nr = nr shr 1
    }
    return res
}
fun modinv(a: Long, p: Long = MOD) : Long = modpow(a, p - 2, p)
fun ncr(n: Long, r: Long) : Long {
    var a = 1L
    var b = 1L
    for (i in 1..r) {
        a = a * (n + 1 - i) % MOD
        b = b * i % MOD
    }
    return modinv(b, MOD) * a % MOD
}

data class Node(val id: Int, var past: Int = -1, var edges: MutableList<Edge> = mutableListOf())
data class Edge(val from: Int, val to: Int, val cost: Long = 1L)