题干：

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 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) = {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 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.

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 integers c1, 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 ≤ n; ai ≠ 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.

Examples

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

题目大意：

   定义点集V(k)和基数Q(k)，分别代表涂有颜色k的点集；颜色为k的所有顶点相连的边的颜色共有多少种（就是 不以顶点为单位，而以颜色为k  为单位）。

   输入n和m，代表n个点m条边，然后输入n个点的颜色，然后输入m条边。（但是一直不知道Output里面那个Node提示有啥用、、）

解题报告：

    set爆搞。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 2e5 + 5;
int n,m;
int col[MAX]; 
set<int> ss[MAX];
int main()
{
	cin>>n>>m;
	for(int i = 1; i<=n; i++) {
		scanf("%d",col+i);
	}
	int u,v;
	for(int i = 1; i<=m; i++) {
		scanf("%d%d",&u,&v);
		if(col[u] != col[v]) {
			ss[col[u]].insert(col[v]);
			ss[col[v]].insert(col[u]);
		}
	}
	int ans = 0,ansi=0x3f3f3f3f;
	for(int i = 1; i<=n; i++) {
		
		if(ss[col[i]].size() > ans) {
			ansi = col[i];
			ans = ss[col[i]].size();			
		}
		else if(ss[col[i]].size() == ans) {
			ansi = min(ansi,col[i]);
			ans = ss[col[i]].size();
		}
//		if(ss[col[i]].size()>= ans) {
//			ansi = min(ansi,col[i]);
//			ans = ss[col[i]].size();
//		}
	}
	printf("%d\n",ansi);
	return 0 ;
 }

总结：

   注意一下最后判断的时候别写成注释那样的，，，，应该跟最短路条数那种判断一样。。。反正注意一下就好了。。