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概述

 D. Colorful Graph

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 ci.

Let's consider all vertices of the graph, that are painted some colork. Let's denote a set of such as V(k). Let's denote the value of theneighbouring color diversity for color k as the cardinality of the set Q(k) = {cu :  cu ≠ 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 setQ(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.

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105) — the number of vertices end edges of the graph, correspondingly. The second line contains a sequence of integersc1, c2, ..., cn (1 ≤ ci ≤ 105) — 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 ai, bi (1 ≤ ai, bi ≤ nai ≠ bi) — 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.

Output

Print the number of the color which has the set of neighbours with the maximum cardinality. It there are multiple optimal colors, print the color with the minimum number. Please note, that you want to find such color, that the graph has at least one vertex with such color.

Example
Input
6 6
1 1 2 3 5 8
1 2
3 2
1 4
4 3
4 5
4 6
Output
3
Input
5 6
4 2 5 2 4
1 2
2 3
3 1
5 3
5 4
3 4
Output
2

n个顶点,各有一个颜色,可以相同,m条边,问相邻的颜色最多的颜色号码是多少

code:

#include <iostream>
#include<cstdio>
#include<set>
#include<cstring>
using namespace std;
set<int>s[100010];
int co[100010];
bool vis[100010];
int main()
{
int n,m,x,y;
scanf("%d%d",&n,&m);
memset(vis,0,sizeof(vis));
for(int i=1;i<=n;i++)
{
s[i].clear();
scanf("%d",&co[i]);
vis[co[i]]=1;
}
for(int i=1;i<=m;i++)
{
scanf("%d%d",&x,&y);
{
if(co[x]!=co[y])
{
s[co[x]].insert(co[y]);
s[co[y]].insert(co[x]);
}
}
}
int maxn=-1,ans;
for(int i=1;i<100001;i++)
{
if(vis[i]&&maxn<(int)s[i].size())
{
maxn=(int)s[i].size();
ans=i;
}
}
printf("%dn",ans);
return 0;
}


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