Problem:
Score : 1000 points

Problem Statement

There is a simple undirected graph with N vertices and M edges.
The vertices are numbered 1 through N, and the edges are numbered 1 through M.
Edge i connects Vertex U_i and V_i.
Also, Vertex i has two predetermined integers A_i and B_i.
You will play the following game on this graph.

First, choose one vertex and stand on it, with W yen (the currency of Japan) in your pocket.
Here, A_s \leq W must hold, where s is the vertex you choose.
Then, perform the following two kinds of operations any number of times in any order:

Choose one vertex v that is directly connected by an edge to the vertex you are standing on, and move to vertex v. Here, you need to have at least A_v yen in your pocket when you perform this move.

Donate B_v yen to the vertex v you are standing on. Here, the amount of money in your pocket must not become less than 0 yen.

You win the game when you donate once to every vertex.
Find the smallest initial amount of money W that enables you to win the game.

Constraints

1 \leq N \leq 10^5

N-1 \leq M \leq 10^5

1 \leq A_i,B_i \leq 10^9

1 \leq U_i < V_i \leq N

The given graph is connected and simple (there is at most one edge between any pair of vertices).

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
:
A_N B_N
U_1 V_1
U_2 V_2
:
U_M V_M

Output

Print the smallest initial amount of money W that enables you to win the game.

Sample Input 1

4 5
3 1
1 2
4 1
6 2
1 2
2 3
2 4
1 4
3 4

Sample Output 1

6

If you have 6 yen initially, you can win the game as follows:

Stand on Vertex 4. This is possible since you have not less than 6 yen.

Donate 2 yen to Vertex 4. Now you have 4 yen.

Move to Vertex 3. This is possible since you have not less than 4 yen.

Donate 1 yen to Vertex 3. Now you have 3 yen.

Move to Vertex 2. This is possible since you have not less than 1 yen.

Move to Vertex 1. This is possible since you have not less than 3 yen.

Donate 1 yen to Vertex 1. Now you have 2 yen.

Move to Vertex 2. This is possible since you have not less than 1 yen.

Donate 2 yen to Vertex 2. Now you have 0 yen.

If you have less than 6 yen initially, you cannot win the game. Thus, the answer is 6.

Sample Input 2

5 8
6 4
15 13
15 19
15 1
20 7
1 3
1 4
1 5
2 3
2 4
2 5
3 5
4 5

Sample Output 2

44

Sample Input 3

9 10
131 2
98 79
242 32
231 38
382 82
224 22
140 88
209 70
164 64
6 8
1 6
1 4
1 3
4 7
4 9
3 7
3 9
5 9
2 5

Sample Output 3

582

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

//Zory-2019
#include<cmath>
#include<ctime>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<map>
#include<set>
#include<queue>
#include<deque>
#include<stack>
#include<bitset>
#include<vector>
#include<algorithm>
#include<iostream>
#include<deque>
using namespace std;
namespace mine
{
    typedef long long ll;
    #define double long double
    const int INF=0x3f3f3f3f;
    const ll LLINF=0x3f3f3f3f3f3f3f3fll;
    ll qread()
    {
        ll ans=0;char c=getchar();int f=1;
        while(c<'0' or c>'9') {if(c=='-') f=-1;c=getchar();}
        while('0'<=c and c<='9') ans=ans*10+c-'0',c=getchar();
        return ans*f;
    }
    void write(ll num)
    {
        if(num<0) {num=-num;putchar('-');}
        if(num>9) write(num/10);
        putchar('0'+num%10);
    }
    void write1(ll num){write(num);putchar(' ');}
    void write2(ll num){write(num);puts("");}
    #define FR first
    #define SE second
    #define MP make_pair
    #define pr pair<int,int>
    #define PB push_back
    #define vc vector
    void chmax(ll &x,const ll y) {x=x>y?x:y;}
    void chmin(ll &x,const ll y) {x=x<y?x:y;}
    const int MAX_N=1e5+10;
    const ll MOD=998244353;
	
	struct DSU
	{
		int fa[MAX_N];DSU(){memset(fa,0,sizeof fa);}
		int findfa(int x){return x==fa[x]?x:findfa(fa[x]);}
	}dsu;
	ll b[MAX_N],c[MAX_N];
	vc<int> son[MAX_N];
	ll g[MAX_N];
	void dp(int x)
	{
		for(int t=0;t<(int)son[x].size();t++)
		{
			int y=son[x][t];dp(y);
			b[x]+=b[y];
		}
		if(son[x].size())
		{
			g[x]=LLINF;
			for(int t=0;t<(int)son[x].size();t++)
			{
				int y=son[x][t];
				chmin(g[x], max(g[y],c[x])+(b[x]-b[y]) );
			}
		}
		else g[x]=b[x]+c[x];
	}
	int pos[MAX_N];bool cmp(int x,int y){return c[x]<c[y];}
	vc<int> to[MAX_N];
    void main()
    {
		int n=qread(),m=qread();for(int i=1;i<=n;i++) c[i]=qread(),b[i]=qread(),c[i]=max(c[i]-b[i],0ll);
		for(int i=1;i<=n;i++) pos[i]=i;sort(pos+1,pos+n+1,cmp);
		for(int i=1;i<=m;i++){int x=qread(),y=qread();to[x].PB(y);to[y].PB(x);}
		for(int i=1;i<=n;i++)
		{
			int x=pos[i];dsu.fa[x]=x;
			for(int t=0;t<(int)to[x].size();t++)
			{
				int y=to[x][t],tt=dsu.findfa(y);
				if(dsu.fa[y]!=0 and tt!=x) dsu.fa[tt]=x,son[x].PB(tt);
			}
		}
		dp(pos[n]);write(g[pos[n]]);
    }
};
signed main()
{
    srand(time(0));
    mine::main();
}