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;
const int N=100005;
int read()
{
   int x=0,f=1;char ch=getchar();
   while (ch<'0'||ch>'9') {if (ch=='-') f=-1;ch=getchar();}
   while (ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
   return x*f;
}

int cnt,head[N],a[N],b[N],id[N],u,v,fa[N],n,m,x,vis[N];
long long f[N],bsum[N];
struct node{int to,next;}num[N*2];
void add(int x,int y) {num[++cnt].to=y;num[cnt].next=head[x];head[x]=cnt;}
bool cmp(int A,int B) {return a[A]<a[B];}
int getfather(int x) {return fa[x]?fa[x]=getfather(fa[x]):x;}
int main()
{
	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),id[i]=i;
	for (int i=1;i<=m;i++) u=read(),v=read(),add(u,v),add(v,u);
	sort(id+1,id+n+1,cmp);
	for (int i=1;i<=n;i++)
	{
		x=id[i];vis[x]=1;f[x]=a[x];bsum[x]=b[x];
		for (int j=head[x];j;j=num[j].next)
		  if (vis[num[j].to])
		  {
		  	  int y=getfather(num[j].to);
		  	  if (x==y) continue;
		  	  fa[y]=x;bsum[x]+=bsum[y];
		  	  f[x]=min(f[x],max(f[y],a[x]-bsum[y]));
		  }
	}
	printf("%lld\n",bsum[id[n]]+f[id[n]]);
    return 0;
}