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)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(register int i = l ; i <= r ; i++)
#define repd(i,r,l) for(register int i = r ; i >= l ; i--)
#define rvc(i,S) for(register int i = 0 ; i < (int)S.size() ; i++)
#define rvcd(i,S) for(register int i = ((int)S.size()) - 1 ; i >= 0 ; i--)
#define fore(i,x)for (register int i = head[x] ; i ; i = e[i].next)
#define forup(i,l,r) for (register int i = l ; i <= r ; i += lowbit(i))
#define fordown(i,id) for (register int i = id ; i ; i -= lowbit(i))
#define pb push_back
#define prev prev_
#define stack stack_
#define mp make_pair
#define fi first
#define se second
#define lowbit(x) (x&(-x))

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int,int> pr;

const ld inf = 2e18;
const int N = 3e6 + 10;
const int maxn = 300020;
const ll mod = 1e9 + 7;
const ull base = 907;

struct node{
	int next,to;
};
node e[maxn * 2];
int head[maxn],cnt;	
int n,m,dfstime;

int a[maxn],b[maxn],vis[maxn],tag[maxn];
int id;
ll sum[maxn],sumf[maxn],f[maxn],mn[maxn];

void clear(){
	rep(i,1,n) head[i] = 0;
	cnt = 0;
}
void adde(int x,int y){
	e[++cnt].to = y;
	e[cnt].next = head[x];
	head[x] = cnt;
}
void init(){
	rep(i,1,m){
		int x,y;
		scanf("%d %d",&x,&y);
		adde(x,y),adde(y,x);
	}
}
void findroot(int x){
	if ( max(0,a[id] - b[id]) < max(0,a[x] - b[x]) ) id = x;
	vis[x] = dfstime;
	fore(i,x){
		if ( vis[e[i].to] == dfstime || tag[e[i].to] ) continue;
		findroot(e[i].to);
	}	
}
ll ans = 0;
int solve(int x){
	++dfstime, id = x , findroot(x);
	int rt = id;
	tag[rt] = 1;

	sum[rt] = b[rt];
	f[rt] = max(a[rt] - b[rt],0);
	fore(i,rt){
		if ( tag[e[i].to] ) continue;
		int to = solve(e[i].to); //注意e[i].to不是下一块的根！
		sum[rt] += sum[to];
		f[rt] = min(f[rt],max(f[to],a[rt] - b[rt] - sum[to]));
	}
	if ( x == 1 ) ans = f[rt] + sum[rt];
	return rt;
}
int main(){
	scanf("%d %d",&n,&m);
	rep(i,1,n) scanf("%d %d",&a[i],&b[i]);
	init();
	solve(1);
	cout<<ans<<endl;
//	rep(i,1,n) cout<<i<<" : "<<f[i]<<" "<<sum[i]<<endl;
}