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.

Examples

Input

4 5
3 1
1 2
4 1
6 2
1 2
2 3
2 4
1 4
3 4


Output

6


Input

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


Output

44


Input

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


Output

582
#include <bits/stdc++.h>
using namespace std;
const int Maxn=100005;
typedef long long ll;
int n,m,f[Maxn],id[Maxn];
ll g[Maxn],sum[Maxn];
vector<int>edge[Maxn],son[Maxn];
struct node{
	int a,b,c;
}w[Maxn];
inline int Getfa(int x){return f[x]=(f[x]==x?x:Getfa(f[x]));}
void dfs1(int x){
	sum[x]=w[x].b;
	for(int y:son[x]){
		dfs1(y);
		sum[x]+=sum[y];
	}
}
void dfs2(int x){
	g[x]=sum[x]+w[x].c;
	for(int y:son[x]){
		dfs2(y);
		g[x]=min(g[x],sum[x]-sum[y]+max(ll(w[x].c),g[y]));
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		scanf("%d%d",&w[i].a,&w[i].b);
		w[i].c=max(w[i].a-w[i].b,0);
		f[i]=id[i]=i;
	}
	for(int i=1,x,y;i<=m;i++){
		scanf("%d%d",&x,&y);
		edge[x].push_back(y);
		edge[y].push_back(x);
	}
	auto cmp=[&](int a,int b){
		return w[a].c<w[b].c||(w[a].c==w[b].c&&a<b);
	};
	sort(id+1,id+n+1,cmp);
	for(int i=1;i<=n;i++)
		for(int y:edge[id[i]])
			if(cmp(Getfa(y),id[i])){
				son[id[i]].push_back(Getfa(y));
				f[Getfa(y)]=id[i];
			}
	dfs1(id[n]);
	dfs2(id[n]);
	cout<<g[id[n]]<<"\n";
	return 0;
}