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>
#define LL long long
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
const int N=100005;
int n,m,fa[N],A[N],B[N],C[N],id[N];bool pd[N];LL sum[N],f[N];
int pp,lnk[N],nxt[N*2],to[N*2];
void ae(int k1,int k2){to[++pp]=k2,nxt[pp]=lnk[k1],lnk[k1]=pp;}
int fd(int k1){return fa[k1]==k1?k1:fa[k1]=fd(fa[k1]);}
bool cmp(int k1,int k2){return C[k1]<C[k2];}
int main(){
	scanf("%d%d",&n,&m);
	rep(i,1,n)scanf("%d%d",&A[i],&B[i]),C[i]=max(A[i]-B[i],0),sum[i]=B[i];
	rep(i,1,m){
		int k1,k2;scanf("%d%d",&k1,&k2);
		ae(k1,k2),ae(k2,k1);
	}
	rep(i,1,n)fa[i]=i,id[i]=i;
	sort(id+1,id+1+n,cmp);
	rep(_,1,n){
		int k1=id[_];
		pd[k1]=1;LL c=C[k1];
		for(int i=lnk[k1];i;i=nxt[i])if(pd[to[i]]){
			int k2=fd(to[i]);if(k2==k1)continue;
			fa[k2]=k1,sum[k1]+=sum[k2],c=min(c,max((LL)C[k1],f[k2])-sum[k2]);
		}
		f[k1]=sum[k1]+c;
//		fprintf(stderr,"f[%d]=%lld\n",k1,f[k1]);
	}
	printf("%lld\n",f[fd(1)]);
	return 0;
}