output_description: Print `YES` if there exists a tree that has the property mentioned by
  Takahashi; print `NO` otherwise.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N M
      a_1 b_1
      :
      a_M b_M
attempt: import sys
  import math
  import copy
  import heapq
  from functools import cmp_to_key
  from bisect import bisect_left, bisect_right
  from collections import defaultdict, deque, Counter
  
  # sys.setrecursionlimit(1000000)
  
  # input aliases
  input = sys.stdin.readline
  getS = lambda: input().strip()
  getN = lambda: int(input())
  getList = lambda: list(map(int, input().split()))
  getZList = lambda: [int(x) - 1 for x in input().split()]
  
  INF = float("inf")
  MOD = 10**9 + 7
  divide = lambda x: pow(x, MOD - 2, MOD)
  
  
  def lcm(x, y):
      return (x * y) // math.gcd(x, y)
  
  
  def nck(n, k, kaijyo):
      return (npk(n, k, kaijyo) * divide(kaijyo[k])) % MOD
  
  
  def npk(n, k, kaijyo):
      if k == 0 or k == n:
          return n % MOD
      return (kaijyo[n] * divide(kaijyo[n - k])) % MOD
  
  
  def fact_and_inv(SIZE):
      inv = [0] * SIZE  # inv[j] = j^{-1} mod MOD
      fac = [0] * SIZE  # fac[j] = j! mod MOD
      finv = [0] * SIZE  # finv[j] = (j!)^{-1} mod MOD
      inv[1] = 1
      fac[0] = fac[1] = 1
      finv[0] = finv[1] = 1
      for i in range(2, SIZE):
          inv[i] = MOD - (MOD // i) * inv[MOD % i] % MOD
          fac[i] = fac[i - 1] * i % MOD
          finv[i] = finv[i - 1] * inv[i] % MOD
  
      return fac, finv
  
  
  def renritsu(A, Y):
      # example 2x + y = 3, x + 3y = 4
      # A = [[2,1], [1,3]])
      # Y = [[3],[4]] または [3,4]
      A = np.matrix(A)
      Y = np.matrix(Y)
      Y = np.reshape(Y, (-1, 1))
      X = np.linalg.solve(A, Y)
  
      # [1.0, 1.0]
      return X.flatten().tolist()[0]
  
  
  class TwoDimGrid:
      # 2次元座標 -> 1次元
      def __init__(self, h, w, wall="#"):
          self.h = h
          self.w = w
          self.size = (h + 2) * (w + 2)
          self.wall = wall
          self.get_grid()
          # self.init_cost()
  
      def get_grid(self):
          grid = [self.wall * (self.w + 2)]
          for i in range(self.h):
              grid.append(self.wall + getS() + self.wall)
  
          grid.append(self.wall * (self.w + 2))
          self.grid = grid
  
      def init_cost(self):
          self.cost = [INF] * self.size
  
      def pos(self, x, y):
          # 壁も含めて0-indexed 元々の座標だけ考えると1-indexed
          return y * (self.w + 2) + x
  
      def getgrid(self, x, y):
          return self.grid[y][x]
  
      def get(self, x, y):
          return self.cost[self.pos(x, y)]
  
      def set(self, x, y, v):
          self.cost[self.pos(x, y)] = v
          return
  
      def show(self):
          for i in range(self.h + 2):
              print(self.cost[(self.w + 2) * i : (self.w + 2) * (i + 1)])
  
      def showsome(self, tgt):
          for t in tgt:
              print(t)
          return
  
      def showsomejoin(self, tgt):
          for t in tgt:
              print("".join(t))
          return
  
      def search(self):
          grid = self.grid
          move = [(0, 1), (0, -1), (1, 0), (-1, 0)]
          move_eight = [
              (0, 1),
              (0, -1),
              (1, 0),
              (-1, 0),
              (1, 1),
              (1, -1),
              (-1, 1),
              (-1, -1),
          ]
          # for i in range(1, self.h+1):
          #     for j in range(1, self.w+1):
          #         cx, cy = j, i
          #         for dx, dy in move_eight:
          #             nx, ny = dx + cx, dy + cy
  
  
  def fprint(c):
      print(c)
      sys.stdout.flush()
  
  
  def solve():
      n, m = getList()
      g = [[] for i in range(n + 1)]
      ji = [0 for i in range(n + 1)]
  
      for i in range(m):
          a, b = getList()
          g[a].append(b)
          g[b].append(a)
          ji[a] += 1
          ji[b] += 1
      # print(ji)
      # print(g)
      for i in range(n - 1):
          if ji[i] % 2 == 1:
              for q in g[i]:
                  if ji[q] % 2 == 0:
                      print("NO")
                      return
  
      print("YES")
  
  
  def main():
      n = getN()
      for _ in range(n):
          solve()
  
      return
  
  
  if __name__ == "__main__":
      # main()
      solve()
  
problem_description: 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.
samples: [{"input": "4 4\n    1 2\n    2 4\n    1 3\n    3 4", "output": "YES\n    \n\nFor example, Takahashi's graph has the property mentioned by him if it has the\nfollowing edges: 1-2, 1-3 and 1-4. In this case, the number written at every\nedge will become 2.\n\n* * *"}, {"input": "5 5\n    1 2\n    3 5\n    5 1\n    3 4\n    2 3", "output": "NO"}]