Toad Ivan has m pairs of integers, each integer is between 1 and n, inclusive. The pairs are (a_1, b_1), (a_2, b_2), …, (a_m, b_m). 

He asks you to check if there exist two integers x and y (1 ≤ x < y ≤ n) such that in each given pair at least one integer is equal to x or y.

Input

The first line contains two space-separated integers n and m (2 ≤ n ≤ 300 000, 1 ≤ m ≤ 300 000) — the upper bound on the values of integers in the pairs, and the number of given pairs.

The next m lines contain two integers each, the i-th of them contains two space-separated integers a_i and b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — the integers in the i-th pair.

Output

Output "YES" if there exist two integers x and y (1 ≤ x < y ≤ n) such that in each given pair at least one integer is equal to x or y. Otherwise, print "NO". You can print each letter in any case (upper or lower).

Examples

Input


4 6
1 2
1 3
1 4
2 3
2 4
3 4


Output


NO


Input


5 4
1 2
2 3
3 4
4 5


Output


YES


Input


300000 5
1 2
1 2
1 2
1 2
1 2


Output


YES

Note

In the first example, you can't choose any x, y because for each such pair you can find a given pair where both numbers are different from chosen integers.

In the second example, you can choose x=2 and y=4.

In the third example, you can choose x=1 and y=2.
'''input
300000 5
1 2
1 2
1 2
1 2
1 2
'''
# its just a simulation
from sys import stdin, setrecursionlimit
import heapq

def get_count(num, old_flag):
	flag = [0] * len(pairs)
	count = 0
	for i in range(len(pairs)):
		if num in pairs[i] and old_flag[i] == 0:
			flag[i] = 1
			count += 1

	return count, flag


def no_interest(num, flag):
	for i in range(len(pairs)):
		if num not in pairs[i] and flag[i] == 0:
			return i
	return -1


setrecursionlimit(15000)


# main starts
n, m = list(map(int, stdin.readline().split()))
pairs = []
for _ in range(m):
	pairs.append(list(map(int, stdin.readline().split())))
#print(pairs)
flag = [0] * m
c1, flag = get_count(pairs[0][0], flag)
index  = no_interest(pairs[0][0], flag)
if index == -1:
	print("YES")
	exit()
c2, flag2 = get_count(pairs[index][0], flag)
c3, flag2 = get_count(pairs[index][1], flag)
# print(c1, c2, c3)
if c1 + c2 == m or c1 + c3 >= m:
	print("YES")
else:
	flag = [0] * m
	c1, flag = get_count(pairs[0][1], flag)
	index  = no_interest(pairs[0][1], flag)
	if index == -1:
		print("YES")
		exit()
	c2, flag2 = get_count(pairs[index][0],flag)
	c3, flag2 = get_count(pairs[index][1], flag)
	if c1 + c2 == m or c1 + c3 == m:
		print("YES")
	else:
		print("NO")



