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;
int n,m,par[100005],fa[100005];
long long a[100005],b[100005],c[100005],s[100005],ans[100005];
vector<int>E[100005],E1[100005];
bool vis[100005];
bool cmp(int x,int y){
	return c[x]<c[y];
}
int find(int x){
	if(x!=fa[x])fa[x]=find(fa[x]);
	return fa[x];
}
void dfs(int u){
	s[u]=b[u];ans[u]=1e18;
	for(int i=0;i<E1[u].size();i++){
		int v=E1[u][i];
		dfs(v);
		s[u]+=s[v];
	}
	for(int i=0;i<E1[u].size();i++){
		int v=E1[u][i];
		ans[u]=min(ans[u],s[u]-s[v]+max(c[u],ans[v]));
	} 
	if(ans[u]==1e18)ans[u]=b[u]+c[u];
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		c[i]=max(0LL,a[i]-b[i]);
		par[i]=fa[i]=i;
	}
	sort(par+1,par+n+1,cmp);
	while(m--){
		int u,v;
		scanf("%d%d",&u,&v);
		E[u].push_back(v);	
		E[v].push_back(u);
	}
	for(int i=1;i<=n;i++){
		int u=par[i];
		for(int j=0;j<E[u].size();j++){
			int v=E[u][j];
			if(vis[v]&&find(v)!=u){
				E1[u].push_back(find(v));
				fa[find(v)]=u;
			}
		}
		vis[u]=1;
	}
	dfs(par[n]);
	printf("%lld\n",ans[par[n]]);
}