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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#define LL long long
#define LD long double
using namespace std;
const int NN=100000 +117;
const int MM= +117;
int read(){
	int fl=1,x;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if(c=='-'){fl=-1;c=getchar();}
	for(x=0;c>='0'&&c<='9';c=getchar())
		x=(x<<3)+(x<<1)+c-'0';
	return x*fl;
}
void open(){
	freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
}
void close(){
	fclose(stdin);
	fclose(stdout);
}


int m,n;
int a[NN]={};
int b[NN]={};
vector<int> p[NN];
bool has[NN]={};
int fa[NN]={};
int ord[NN]={};
LL dp[NN]={};
LL sumb[NN]={};
int getfa(int x){
	return x==fa[x]?x:fa[x]=getfa(fa[x]);
}
bool cmp(int x,int y){
	return a[x]<a[y];
}
int main(){
	//open();
	n=read();
	m=read();
	for(int i=1;i<=n;++i){
		a[i]=read();
		b[i]=read();
		a[i]=max(a[i]-b[i],0);
		fa[i]=i;
		ord[i]=i;
	}
	for(int i=1;i<=m;++i){
		int x=read(),y=read();
		p[x].push_back(y);
		p[y].push_back(x);
	}
	sort(ord+1,ord+1+n,cmp);
	for(int i=1;i<=n;++i){
		int x=ord[i];
		has[x]=1;
		dp[x]=a[x];
		sumb[x]=b[x];
		for(int j=0;j<p[x].size();++j){
			int cur=p[x][j];
			if(has[cur]){
				int rt=getfa(cur);
				if(rt==x)continue;
				sumb[x]+=sumb[rt];
				dp[x]=min(dp[x],max(dp[rt],a[x]-sumb[rt]));
				fa[rt]=x;
			}
		}

	}
	printf("%lld\n",dp[getfa(1)]+sumb[getfa(1)]);
	close();
	return 0;
}