You've got an undirected graph, consisting of *n* vertices and *m* edges. We will consider the graph's vertices numbered with integers from 1 to *n*. Each vertex of the graph has a color. The color of the *i*-th vertex is an integer *c*<sub class="lower-index">*i*</sub>.

Let's consider all vertices of the graph, that are painted some color *k*. Let's denote a set of such as *V*(*k*). Let's denote the value of the neighbouring color diversity for color *k* as the cardinality of the set *Q*(*k*) = {*c*<sub class="lower-index">*u*</sub> :  *c*<sub class="lower-index">*u*</sub> ≠ *k* and there is vertex *v* belonging to set *V*(*k*) such that nodes *v* and *u* are connected by an edge of the graph}.

Your task is to find such color *k*, which makes the cardinality of set *Q*(*k*) maximum. In other words, you want to find the color that has the most diverse neighbours. Please note, that you want to find such color *k*, that the graph has at least one vertex with such color.
The first line contains two space-separated integers *n*, *m* (1 ≤ *n*, *m* ≤ 10<sup class="upper-index">5</sup>) — the number of vertices end edges of the graph, correspondingly. The second line contains a sequence of integers *c*<sub class="lower-index">1</sub>, *c*<sub class="lower-index">2</sub>, ..., *c*<sub class="lower-index">*n*</sub> (1 ≤ *c*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>) — the colors of the graph vertices. The numbers on the line are separated by spaces.

Next *m* lines contain the description of the edges: the *i*-th line contains two space-separated integers *a*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*i*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub>, *b*<sub class="lower-index">*i*</sub> ≤ *n*; *a*<sub class="lower-index">*i*</sub> ≠ *b*<sub class="lower-index">*i*</sub>) — the numbers of the vertices, connected by the *i*-th edge. 

It is guaranteed that the given graph has no self-loops or multiple edges.
#!/usr/bin/python3
# D. Colorful Graph
from collections import deque

MAX_COLORS = 10 

def bfs(vertex):
	global adjlist, visited, colors, dvt, count, idx
	queue = deque()
	queue.append(vertex)
	visited[vertex] = 1
	while queue:
		curv = queue.popleft()
		for v in adjlist[curv]:
			if colors[v] != colors[curv]:
				if colors[curv] not in dvt:
					dvt[colors[curv]] = {colors[v]}
				else:
					dvt[colors[curv]].add(colors[v])
				if len(dvt[colors[curv]]) > count:
					idx = colors[curv]
					count = len(dvt[colors[curv]])
				elif len(dvt[colors[curv]]) == count and colors[curv] < idx:
					idx = colors[curv]
			if visited[v] == 0:
				queue.append(v)
				visited[v] = 1

n, m = [int(x) for x in input().split()]
colors = []
adjlist = [[] for x in range(n)]
dvt = {} 
visited = [0] * n
colors = [int(x) for x in input().split()]
for i in range(m):
	x, y = [int(x) for x in input().split()]
	x -= 1
	y -= 1
	adjlist[x].append(y)
	adjlist[y].append(x)

count = 0
idx = 0
for i in range(n):
	if visited[i] == 0:
		bfs(i)

print(idx)
