題幹：

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

題目大意：

   定義點集V(k)和基數Q(k)，分別代表塗有顏色k的點集；顏色為k的所有頂點相連的邊的顏色共有多少種（就是 不以頂點為單位，而以顏色為k  為單位）。

   輸入n和m，代表n個點m條邊，然後輸入n個點的顏色，然後輸入m條邊。（但是一直不知道Output裡面那個Node提示有啥用、、）

解題報告：

    set爆搞。

AC程式碼：

#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 ;
 }

總結：

   注意一下最後判斷的時候別寫成註釋那樣的，，，，應該跟最短路條數那種判斷一樣。。。反正注意一下就好了。。