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<cmath>
#include<algorithm>
#include<cstdlib>
#include<cstring>
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
int stack[20];
inline void write(int x)
{
	if(x<0){putchar('-');x=-x;}
	if(!x){putchar('0');return;}
	int top=0;
	while(x)stack[++top]=x%10,x/=10;
	while(top)putchar(stack[top--]+'0');
}
inline void pr1(int x){write(x),putchar(' ');}
inline void pr2(int x){write(x),puts("");}
struct node
{
	int x,y,next;
}e[100010];int len,last[100010];
inline void ins(int x,int y)
{
	len++;
	e[len].x=x;e[len].y=y;
	e[len].next=last[x];last[x]=len;
}
long long a[100010],b[100010];
bool cmp(node u,node v){return a[u.x]<a[v.x];}
int rt,fa[100010];
inline int findfa(int x)
{
	if(fa[x]!=x)fa[x]=findfa(fa[x]);
	return fa[x];
}
long long uf,tot[100010],s[100010];
inline void dfs(int x)
{
	for(int k=last[x];k;k=e[k].next)
	{
		int y=e[k].y;
		s[y]=max(s[x],a[x]-tot[y]);
		dfs(y);
	}
}
int main()
{
    //freopen("a.in","r",stdin);
    //freopen("a.out","w",stdout);
    int n=read(),m=read();
    for(int i=1;i<=n;i++){a[i]=read(),b[i]=read();a[i]=max(0LL,a[i]-b[i]),uf+=b[i];}
    for(int i=1;i<=m;i++)
    {
    	int x=read(),y=read();if(a[x]<a[y])swap(x,y);
    	e[i].x=x,e[i].y=y;
    }sort(e+1,e+m+1,cmp);
    for(int i=1;i<=n;i++)fa[i]=i,tot[i]=b[i];
    for(int i=1;i<=m;i++)
    {
    	int fx=findfa(e[i].x),fy=findfa(e[i].y);
    	if(fx!=fy){fa[fy]=fx,rt=fx,tot[fx]+=tot[fy];ins(fx,fy);}
    }dfs(rt);
    long long ans=max(a[1],s[1])+uf;
    for(int i=2;i<=n;i++)ans=min(ans,max(a[i],s[i])+uf);
    printf("%lld\n",ans);
    return 0;
}