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 <stdio.h>
#include <string.h>
#include <algorithm>
#include <vector>

#define rep(i,st,ed) for (int i=st;i<=ed;++i)

typedef long long LL;
const LL INF=1e18;
const int N=200005;

std:: vector <int> E[N],G[N];

LL a[N],b[N],f[N],g[N];
int fa[N],rk[N];

bool vis[N];

int read() {
	int x=0,v=1; char ch=getchar();
	for (;ch<'0'||ch>'9';v=(ch=='-')?(-1):v,ch=getchar());
	for (;ch<='9'&&ch>='0';x=x*10+ch-'0',ch=getchar());
	return x*v;
}

void dfs(int x) {
	if (G[x].empty()) {
		f[x]=a[x]+b[x],g[x]=b[x];
		return ;
	}
	g[x]=b[x],f[x]=INF;
	for (int y:G[x]) dfs(y),g[x]+=g[y];
	for (int y:G[x]) {
		f[x]=std:: min(f[x],std:: max(f[y],a[x])+g[x]-g[y]);
	}
}

int find(int x) {
	return !fa[x]?x:(fa[x]=find(fa[x]));
}

bool cmp(int x,int y) {
	return a[x]<a[y];
}

int main(void) {
	int n=read(),m=read();
	rep(i,1,n) {
		a[i]=read(),b[i]=read();
		a[i]=std:: max(a[i]-b[i],0LL);
		rk[i]=i;
	}
	rep(i,1,m) {
		int x=read(),y=read();
		E[x].push_back(y);
		E[y].push_back(x);
	}
	std:: sort(rk+1,rk+n+1,cmp);
	rep(ti,1,n) {
		int x=rk[ti]; vis[x]=1;
		for (int y:E[x]) {
			if (!vis[y]) continue;
			if (find(y)==find(x)) continue;
			G[x].push_back(find(y));
			fa[find(y)]=find(x);
		}
	}
	rep(i,1,n) if (find(i)==i) {
		dfs(i),printf("%lld\n", f[i]);
	}
	return 0;
}