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<algorithm>
#include<cstdio>
#define mxn 100010
#define LL long long
using namespace std;
int n,m,sl,fh,p[mxn],fa[mxn],vis[mxn];
LL ans,a[mxn],b[mxn],s[mxn];
int t,h[mxn];
struct edge
{
	int to,nxt;
}e[mxn<<1];
int rd()
{
	sl=0;fh=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch) {if(ch=='-') fh=-1; ch=getchar();}
	while('0'<=ch&&ch<='9') sl=sl*10+ch-'0',ch=getchar();
	return sl*fh;
}
bool cmp(const int &x,const int &y) {return s[x]<s[y];}
void add(int u,int v)
{
	e[++t]=(edge){v,h[u]};h[u]=t;
	e[++t]=(edge){u,h[v]};h[v]=t;
}
int getfa(int x)
{
	if(fa[x]!=x) fa[x]=getfa(fa[x]);
	return fa[x];
}
int main()
{
	n=rd();m=rd();int x,y;
	for(int i=1;i<=n;++i)
	{
		a[i]=rd();b[i]=rd();
		s[i]=a[i]=max(0ll,a[i]-b[i]);
		fa[i]=p[i]=i;
	}
	sort(p+1,p+n+1,cmp);
	for(int i=1;i<=m;++i) x=rd(),y=rd(),add(x,y);
	int u,v,fx,fy;
	for(int k=1;k<=n;++k)
	{
		vis[u=p[k]]=1;
		for(int i=h[u];i;i=e[i].nxt)
			if(vis[v=e[i].to])
			{
				fx=getfa(u);fy=getfa(v);
				if(fx==fy) continue;
				b[fx]+=b[fy];fa[fy]=fx;
				s[fx]=min(s[fx],s[fy]+max(0ll,a[fx]-s[fy]-b[fy]));
			}
	}
	printf("%lld\n",s[p[n]]+b[p[n]]);
	return 0;
}