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>
const int N=100005;
using namespace std;
long long b[N],g[N];
int n,m,a[N],c[N],no[N],re[N],p[N],f[N],
to[N*2],hd[N*2],lk[N],cnt;
bool vis[N];
inline void add(int u,int v){
	to[++cnt]=v,hd[cnt]=lk[u],lk[u]=cnt;
}int u,v;
int get(int x){return p[x]?p[x]=get(p[x]):x;}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
	scanf("%d%lld",a+i,c+i),
	no[i]=i,a[i]-=c[i];
	sort(no,no+n,[](int x,int y){
		return a[x]<a[y];
	});
	sort(a,a+n);
	for(int i=0;i<n;i++)
	b[i]=c[no[i]],re[no[i]]=i,g[i]=max(a[i],0);
	for(;m--;)
	scanf("%d%d",&u,&v),
	u=re[u-1],v=re[v-1],
	add(u,v),add(v,u);
	for(int i=0,j,s;i<n;i++){
		vis[i]=1;
		for(j=lk[i];j;j=hd[j])
		if(vis[s=to[j]]){
			if((u=get(s))<i)
			f[u]=p[u]=i,g[i]=1e18;
		}
	}
	for(int i=0;i<n;i++){
		if(i<n-1)b[f[i]]+=b[i];
		g[f[i]]=min(g[f[i]],max(a[f[i]]-b[i],g[i]));
	}
	printf("%lld",g[n-1]+b[n-1]);
}