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 <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N=100010;
const LL LINF=1ll<<60;
int n,m;
int cost[N];
int h[N],tot;
struct Edge{
	int v,next;
}e[N*2];
struct Node{
	int u,t;
}p[N];
inline LL min(LL x,LL y){
	return x<y?x:y;
}
inline LL max(LL x,LL y){
	return x>y?x:y;
}
void addEdge(int u,int v){
	e[++tot]=(Edge){v,h[u]}; h[u]=tot;
	e[++tot]=(Edge){u,h[v]}; h[v]=tot;
}
void readData(){
	scanf("%d%d",&n,&m);
	int a,b;
	for(int i=1;i<=n;i++){
		scanf("%d%d",&a,&b);
		cost[i]=b;
		p[i]=(Node){i,max(a-b,0)};
	}
	int u,v;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&u,&v);
		addEdge(u,v);
	}
}
int bel[N];
bool active[N];
LL f[N],sum[N];
bool cmpThreshold(const Node &x,const Node &y){
	return x.t<y.t;
}
int find(int x){
	return bel[x]==x?x:(bel[x]=find(bel[x]));
}
void solve(){
	sort(p+1,p+1+n,cmpThreshold);
	for(int i=1;i<=n;i++) bel[i]=i;
	for(int ii=1;ii<=n;ii++){
		int u=p[ii].u;
		active[u]=true;
		sum[u]=cost[u];
		f[u]=LINF;
		bool haveSon=false;
		for(int i=h[u],v;i;i=e[i].next){
			v=find(e[i].v);
			if(v!=u&&active[v]){
				active[v]=false;
				haveSon=true;
				bel[v]=u;
				sum[u]+=sum[v];
				f[u]=min(f[u],max(f[v],p[ii].t-sum[v]));
			}
		}
		if(!haveSon)
			f[u]=p[ii].t;
	}
	int last=p[n].u;
	printf("%lld\n",sum[last]+f[last]);
}
int main(){
	readData();
	solve();
	return 0;
}
