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<bits/stdc++.h>
#define ll long long
#define dbg1(x) cerr<<#x<<"="<<(x)<<" "
#define dbg2(x) cerr<<#x<<"="<<(x)<<"\n"
#define dbg3(x) cerr<<#x<<"\n"
using namespace std;
#define reg register
#define pb push_back
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<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
const int MN=1e5+5;
int N,M,id[MN],rk[MN],par[MN];
ll f[MN],sz[MN],A[MN],B[MN],C[MN];
vector<int> G[MN],T[MN];
bool cmp(const int&x,const int&y){return C[x]<C[y];}
int getf(int x){return x==par[x]?x:par[x]=getf(par[x]);}
void rw(ll &x,ll y){if(y<x)x=y;}
void dfs(int x)
{
	#define y T[x][i]
	reg int i;sz[x]=B[x];f[x]=1e15;
	if(!T[x].size())return (void)(f[x]=B[x]+C[x]);
	for(i=T[x].size()-1;~i;--i)dfs(y),sz[x]+=sz[y];
	for(i=T[x].size()-1;~i;--i)rw(f[x],sz[x]-sz[y]+max(C[x],f[y]));
	#undef y
}
int main()
{
	N=read();M=read();
	reg int i,j,x,y;
	for(i=1;i<=N;++i) 
        par[i]=id[i]=i;
	for(i=1;i<=N;++i){ 
        A[i]=read();
        B[i]=read();
        C[i]=max(A[i]-B[i],0LL);
    }
	for(i=1;i<=M;++i) {
        x=read();
        y=read();
        G[x].pb(y);
        G[y].pb(x);
    }
	sort(id+1,id+N+1,cmp);
	for(i=1;i<=N;++i)rk[id[i]]=i;
	for(i=1;i<=N;++i)for(j=G[x=id[i]].size()-1;~j;--j){
		y=getf(G[x][j]);
		if(y!=x&&rk[y]<i) 
            par[y]=x,T[x].pb(y);
	}
	dfs(id[N]);
	return 0*printf("%lld\n",f[id[N]]);
}