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;
typedef long long ll;
const int N=100005;
int n,m,ed,a[N],fir[N],vis[N],b[N],id[N],c[N],f[N];
ll sum[N],g[N];
struct E{int v,nxt;}e[N*2];  
void adde(int x,int y){
	e[++ed].v=y;e[ed].nxt=fir[x];fir[x]=ed;
}
bool cmp(int x,int y){
	return c[x]<c[y]; 
} 
int F(int x){return x==f[x]?x:f[x]=F(f[x]);}
int main(){
	//freopen("aa.in","r",stdin);
	int i,j;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;++i){
		scanf("%d%d",a+i,b+i);
		c[i]=max(a[i]-b[i],0);	
	}
	for(i=1;i<=m;++i){
		int x,y;scanf("%d%d",&x,&y);
		adde(x,y);adde(y,x);
	}
	for(i=1;i<=n;++i)f[i]=id[i]=i,g[i]=c[i],sum[i]=b[i];
	sort(id+1,id+1+n,cmp);
	for(i=1;i<=n;++i){
		int x=id[i];vis[x]=1;
		for(j=fir[x];j;j=e[j].nxt)if(vis[e[j].v]){
			int y=e[j].v;
			if(x==F(y))continue;
			//printf("%d %d\n",x,y);
			//printf("con %d %d ,f %d %d : %d %d\n",x,y,f[x],f[y],sum[f[x]],sum[f[y]]);
			g[x]=min(g[x],g[f[y]]+max(c[x]-g[f[y]]-sum[f[y]],0LL));
			//printf("g %lld\n",g[f[x]]);		
			sum[x]+=sum[f[y]];
			f[f[y]]=x;
		}
	} 
	//printf("fin sum %lld\n",sum[F(1)]); 
	ll ans=g[F(1)]+sum[F(1)];
	cout<<ans<<endl;
	return 0;
} 