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 <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=1e5+5;
typedef long long ll;
ll dp[maxn],s[maxn];
ll val[maxn];
vector<int> adj[maxn],bdj[maxn];
int a[maxn],b[maxn],n,m;
pair<int,int> c[maxn];
int ufs[maxn];
int vis[maxn];
inline int find(int u){
	if (ufs[u]==u) return u;
	else return ufs[u]=find(ufs[u]);
}
inline bool cmp1(pair<int,int> a1,pair<int,int> b1){
	return a1.first<b1.first;
}
inline void dfs(int u,int fa){
	s[u]=b[u];
	for (int i=0;i<bdj[u].size();i++){
		int v=bdj[u][i];
		if (v==fa) continue;
		dfs(v,u);
		s[u]+=s[v];
	}
	if (bdj[u].size()==0){
		dp[u]=max(a[u],b[u]);
		return;
	}
	else dp[u]=1e16;
	for (int i=0;i<bdj[u].size();i++){
		int v=bdj[u][i];
		if (v==fa) continue;
		dp[u]=min(dp[u],max(dp[v],val[u])+s[u]-s[v]);
	}
}
int main(){
	cin>>n>>m;
	for (int i=1;i<=n;i++){
		scanf("%d%d",&a[i],&b[i]);
		c[i].first=max(0,a[i]-b[i]); 
		val[i]=c[i].first;
		c[i].second=i;
		ufs[i]=i;
	}
	for (int i=1;i<=m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	sort(c+1,c+n+1,cmp1);
	for (int i=1;i<=n;i++){
		int id=c[i].second;
		vis[id]=1;
		for (int j=0;j<adj[id].size();j++){
			int v=adj[id][j];
			if (vis[v]==0) continue;
			v=find(v);
			if (v==id) continue;
			bdj[id].push_back(v);
			ufs[v]=id;
		}
	}
	dfs(c[n].second,c[n].second);
	cout<<dp[c[n].second]<<endl;
}