human
Bear Limak examines a social network. Its main functionality is that two members can become friends (then they can talk with each other and share funny pictures).

There are n members, numbered 1 through n. m pairs of members are friends. Of course, a member can't be a friend with themselves.

Let A-B denote that members A and B are friends. Limak thinks that a network is reasonable if and only if the following condition is satisfied: For every three distinct members (X, Y, Z), if X-Y and Y-Z then also X-Z.

For example: if Alan and Bob are friends, and Bob and Ciri are friends, then Alan and Ciri should be friends as well.

Can you help Limak and check if the network is reasonable? Print "YES" or "NO" accordingly, without the quotes.

Input

The first line of the input contain two integers n and m (3 ≤ n ≤ 150 000, ) — the number of members and the number of pairs of members that are friends.

The i-th of the next m lines contains two distinct integers ai and bi (1 ≤ ai, bi ≤ n, ai ≠ bi). Members ai and bi are friends with each other. No pair of members will appear more than once in the input.

Output

If the given network is reasonable, print "YES" in a single line (without the quotes). Otherwise, print "NO" in a single line (without the quotes).

Examples

Input

4 3
1 3
3 4
1 4

Output

YES

Input

4 4
3 1
2 3
3 4
1 2

Output

NO

Input

10 4
4 3
5 10
8 9
1 2

Output

YES

Input

3 2
1 2
2 3

Output

NO

Note

The drawings below show the situation in the first sample (on the left) and in the second sample (on the right). Each edge represents two members that are friends. The answer is "NO" in the second sample because members (2, 3) are friends and members (3, 4) are friends, while members (2, 4) are not.

Using python can you solve the prior task?

=====
Can you provide a Python code to print "YES" or "NO" accordingly, without the quotes?
assistant
#Here is Python code to print "YES" or "NO" accordingly, without the quotes.

"""Template for Python Competitive Programmers prepared by Mayank Chaudhary """

# to use the print and division function of Python3
from __future__ import division, print_function

"""value of mod"""
MOD = 998244353
mod = 10**9 + 7

"""use resource"""
# import resource
# resource.setrlimit(resource.RLIMIT_STACK, [0x100000000, resource.RLIM_INFINITY])

"""for factorial"""

# def prepare_factorial():
#     fact = [1]
#     for i in range(1, 105):
#         fact.append((fact[-1] * i) % mod)
#     ifact = [0] * 105
#     ifact[104] = pow(fact[104], mod - 2, mod)
#     for i in range(104, 0, -1):
#         ifact[i - 1] = (i * ifact[i]) % mod
#
#     return fact, ifact

"""uncomment next 4 lines while doing recursion based question"""
import threading
threading.stack_size(1<<27)
import sys
sys.setrecursionlimit(10000)

"""uncomment modules according to your need"""
# from bisect import bisect_left, bisect_right, insort
# import itertools
# from math import floor, ceil, sqrt, degrees, atan, pi, log
# from heapq import heappop, heapify, heappush
# from random import randint as rn
# from Queue import Queue as Q
from collections import Counter, defaultdict, deque
# from copy import deepcopy
'''
def modinv(n, p):
    return pow(n, p - 2, p)
'''

# def ncr(n, r,  fact, ifact):  # for using this uncomment the lines calculating fact and ifact
#     t = (fact[n] * (ifact[r]*ifact[n-r]) % mod)% mod
#     return t


def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()

# def GCD(x, y):
#     while (y):
#         x, y = y, x % y
#     return x

# def lcm(x, y):
#     return (x*y)//(GCD(x, y))

# def get_xor(n):
#     return [n,1,n+1,0][n%4]

# def binary_expo(a, b):
#
#     result = 1
#     while b:
#         if b&1:
#             result *= a
#             b-=1
#         else:
#             a *= a
#             b >>= 1
#     return result

"""...