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: Accepted)

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;
struct node{
	LL y,next;
}e[500010];LL len=0,last[500010];
void ins(LL x,LL y)
{
	e[++len].y=y;
	e[len].next=last[x];last[x]=len;
}
LL n,m,fa[500010],f[500010],sum[500010];
LL findfa(LL x) {return fa[x]==x?x:fa[x]=findfa(fa[x]);}
LL a[500010],b[500010],pos[500010],vis[500010];
bool cmp(LL x,LL y) {return a[x]<a[y];}
int main()
{
	scanf("%lld %lld",&n,&m);
	for(LL i=1;i<=n;i++)
	{
		scanf("%lld %lld",&a[i],&b[i]);
		a[i]=max(0LL,a[i]-b[i]);
	}
	for(LL i=1;i<=m;i++)
	{
		LL x,y;scanf("%lld %lld",&x,&y);
		ins(x,y);ins(y,x);
	}
	for(LL i=1;i<=n;i++) fa[i]=i,sum[i]=b[i],pos[i]=i,f[i]=a[i];
	sort(pos+1,pos+n+1,cmp);
	memset(vis,false,sizeof(vis));
	for(LL j=1;j<=n;j++)
	{
		LL x=pos[j];vis[x]=1;
		for(LL i=last[x];i;i=e[i].next)
		{
			LL y=e[i].y;
			if(vis[y])
			{
				LL tx=findfa(x),ty=findfa(y);
				if(tx==ty) continue;
				sum[tx]+=sum[ty];fa[ty]=tx;
				f[tx]=min(f[tx],f[ty]+max(0LL,a[x]-f[ty]-sum[ty]));
			}
		}
	}
	printf("%lld\n",sum[pos[n]]+f[pos[n]]);
}