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
/*
  mail: mleautomaton@foxmail.com
  author: MLEAutoMaton
  This Code is made by MLEAutoMaton
*/
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<map>
#include<iostream>
using namespace std;
#define ll long long
#define REP(a,b,c) for(int a=b;a<=c;a++)
#define re register
#define int ll
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
inline int gi(){
	int f=1,sum=0;char ch=getchar();
	while(ch>'9' || ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0' && ch<='9'){sum=(sum<<3)+(sum<<1)+ch-'0';ch=getchar();}
	return f*sum;
}
const int N=200010,Inf=1e18+10;
int n,m,a[N],b[N],sum[N],f[N],p[N],dp[N],vis[N];
int find(int x){return f[x]!=x?f[x]=find(f[x]):f[x];}
vector<int>G[N];
bool cmp(int A,int B){return a[A]<a[B];}
signed main(){
	n=gi();m=gi();
	for(int i=1;i<=n;i++){
		a[i]=gi();b[i]=gi();a[i]=max(a[i]-b[i],0ll);
		p[i]=f[i]=i;sum[i]=b[i];dp[i]=a[i];
	}
	for(int i=1;i<=m;i++){int u=gi(),v=gi();G[u].push_back(v);G[v].push_back(u);}
	sort(p+1,p+n+1,cmp);
	for(int i=1;i<=n;i++){
		int u=p[i];vis[u]=1;
		for(int v:G[u])
			if(vis[v]){
				int x=find(u),y=find(v);
				if(x!=y){
					f[y]=x;sum[x]+=sum[y];
					dp[x]=min(dp[x],max(dp[y],a[x]-sum[y]));
				}
			}
	}
	printf("%lld\n",dp[find(1)]+sum[find(1)]);
	return 0;
}
