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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <algorithm>
#define int long long
#define N 100002
using namespace std;
int head[N],ver[N*2],nxt[N*2],l;
int head1[N],ver1[N*2],nxt1[N*2],l1;
int n,m,i,j,a[N],b[N],c[N],fa[N],p[N],f[N],sum[N];
bool vis[N];
int read()
{
	char c=getchar();
	int w=0;
	while(c<'0'||c>'9') c=getchar();
	while(c<='9'&&c>='0'){
		w=w*10+c-'0';
		c=getchar();
	}
	return w;
}
void insert(int x,int y)
{
	l++;
	ver[l]=y;
	nxt[l]=head[x];
	head[x]=l;
}
void insert1(int x,int y)
{
	l1++;
	ver1[l1]=y;
	nxt1[l1]=head1[x];
	head1[x]=l1;
}
int my_comp(const int &x,const int &y)
{
	return c[x]<c[y];
}
int find(int x)
{
	if(fa[x]!=x) fa[x]=find(fa[x]);
	return fa[x];
}
void dfs(int x)
{
	sum[x]=b[x];
	for(int i=head1[x];i;i=nxt1[i]){
		int y=ver1[i];
		dfs(y);
		sum[x]+=sum[y];
	}
	f[x]=sum[x]+c[x];
	for(int i=head1[x];i;i=nxt1[i]){
		int y=ver1[i];
		f[x]=min(f[x],sum[x]-sum[y]+max(f[y],c[x]));
	}
}
signed main()
{
	n=read();m=read();
	for(i=1;i<=n;i++){
		a[i]=read();b[i]=read();
		c[i]=max(a[i]-b[i],1LL*0);
	}
	for(i=1;i<=m;i++){
		int u=read(),v=read();
		insert(u,v);
		insert(v,u);
	}
	for(i=1;i<=n;i++) fa[i]=p[i]=i;
	sort(p+1,p+n+1,my_comp);
	for(i=1;i<=n;i++){
		int x=p[i];
		vis[x]=1;
		for(j=head[x];j;j=nxt[j]){
			int y=ver[j];
			if(vis[y]){
				int f=find(y);
				if(f!=x){
					fa[f]=x;
					insert1(x,f);
				}
			}
		}
	}
	dfs(p[n]);
	cout<<f[p[n]]<<endl;
	return 0;
}