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>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
const int N = 201000;
ll A[N],B[N],s[N],f[N];
int n,m,p[N],fa[N];
vector<int> e[N],a[N];
int find(int x){return fa[x]= fa[x]==x?x:find(fa[x]);}
void dp(int x){
	s[x]=B[x];
	for(int v:e[x]) dp(v),s[x]+=s[v];
	if(!e[x].size()){
		f[x]=max(A[x],B[x]);
		return;
	}
	f[x]=1e18;
	for(int v:e[x])
		f[x]=min(f[x],max(f[v],A[x]-B[x])+s[x]-s[v]);
}
int main(){
	read(n);read(m);
	for(int i=1;i<=n;i++) read(A[i]),read(B[i]);
	for(int i=1;i<=m;i++){
		int x,y;read(x);read(y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	for(int i=1;i<=n;i++) p[i]=i,fa[i]=i;
	sort(p+1,p+n+1,[&](int a,int b){
		return A[a]-B[a]<A[b]-B[b];
	});
	for(int i=1;i<=n;i++){
		int x=p[i];
		for(int v:a[x])
			if(A[v]-B[v]<=A[x]-B[x]&&find(v)!=find(x)){
				e[x].push_back(find(v));
				fa[find(v)]=find(x);
			}
	}
	for(int i=1;i<=n;i++)
		if(fa[i]==i){
			dp(i);
			cout<<f[i]<<'\n';
			return 0;
		}
	return 0;
}
