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<iostream>
#include<algorithm>
#include<ctime>
#include<cstdlib>
#include<sys/time.h>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long ll;
inline int ri()
{
	register int x=0;register bool f=0;register char ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')f=1;ch=getchar();}
	while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();};
	return f?-x:x;
}
void wi(int x)
{
	if(x<0)	x=-x,putchar('-');
	if(x>9)	wi(x/10);
	putchar('0'+x%10);
}
#define N 100005
struct edge
{
	int to,next;
}e[N*2];
int n,m,tot,head[N];
int a[N],b[N],id[N],fa[N];
long long qd[N],S[N];
bool cmp(int x,int y)
{
	return a[x]<a[y];
}
void add(int x,int y)
{
	e[++tot]=(edge){y,head[x]};
	head[x]=tot;
}
int get(int x)
{
	return fa[x]==x?x:fa[x]=get(fa[x]);
}
int vis[N];
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&a[i],&b[i]);
		a[i]=max(a[i]-b[i],0);
	}
	for(int i=1;i<=m;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y);add(y,x);
	}
	for(int i=1;i<=n;i++)	id[i]=i,fa[i]=i,qd[i]=a[i],S[i]=b[i]; 
	sort(id+1,id+n+1,cmp);
	for(int i=1;i<=n;i++)
	{
		int x=id[i];
		vis[x]=1;
		for(int j=head[x];j;j=e[j].next)
			if(vis[e[j].to])
			{
				int X=get(x),Y=get(e[j].to);
				if(X==Y) continue;
				fa[Y]=X;S[X]+=S[Y];
				qd[X]=min(qd[X],qd[Y]+max(0ll,a[x]-S[Y]-qd[Y]));
			}
	}
	int x=get(1);
	printf("%lld",qd[x]+S[x]);
}