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: Wrong Answer)

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

int n,m,a[100010],b[100010],id[100010],fa[100010],bel[100010],x,y,d[100010],p[100010],q[100010],l,r;
long long sum[100010],f[100010],ans;
vector<int> v[100010];

bool cmp(const int &x,const int &y) {return a[x]<a[y];}
int ask(int x) {return bel[x]==x?x:bel[x]=ask(bel[x]);}
void Addedge(int x,int y) {v[x].push_back(y),v[y].push_back(x);}

int main()
{
	srand(12019687);
	scanf("%d%d",&n,&m),ans=1000000000000000000ll;
	for (int i=1; i<=n; i++) scanf("%d%d",&a[i],&b[i]),id[i]=i;
	for (int i=1; i<=m; i++) scanf("%d%d",&x,&y),Addedge(x,y);
	sort(id+1,id+1+n,cmp);
	for (int qaq=1; qaq<=130; qaq++)
	{
		for (int l=1,r; l<=n; l++)
		{
			r=l;
			while (r<n&&a[id[r+1]]==a[id[l]]) r++;
			random_shuffle(id+l,id+r+1),l=r;
		}
		for (int i=1; i<=n; i++) p[id[i]]=i,d[i]=0;
		for (int i=1; i<=n; i++) bel[i]=i,fa[i]=0;
		for (int i=1; i<=n; i++)
		{
			x=id[i];
			for (int j:v[x])
				if (p[j]<i&&(y=ask(j))!=x) fa[y]=x,bel[y]=x;
		}
		l=1,r=0;
		for (int i=1; i<=n; i++) if (fa[i]) d[i]++,d[fa[i]]++;
		for (int i=1; i<=n; i++) if (d[i]==1&&fa[i]) q[++r]=i;
		while (l<=r)
		{
			x=q[l++];
			if (fa[x]&&(--d[fa[x]])==(fa[x]!=id[n])) q[++r]=fa[x];
		}
		sum[0]=0;
		for (int i=1; i<=n; i++) f[i]=1000000000000000000ll,sum[i]=b[i];
		for (int i=1; i<=r; i++) 
		{
			x=q[i];
			f[x]=min(sum[x]-b[x]+f[x],sum[x]-b[x]+max(a[x],b[x]));
			f[fa[x]]=min(f[fa[x]],max(1ll*a[fa[x]],b[fa[x]]+f[x])-sum[x]);
			sum[fa[x]]+=sum[x];
		}
		ans=min(ans,f[id[n]]);
	}
	printf("%lld\n",ans);
	return 0;
}