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>
typedef long long ll;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;
const int MAX_N=1e5,MAX_M=1e5;

int N,M;
ll A[MAX_N],B[MAX_N],C[MAX_N];
const ll INF=1e17;
vector<int> G[MAX_N];

int tpar[MAX_N];
vector<int> tre[MAX_N];
ll dp[MAX_N],bsm[MAX_N];

int upar[MAX_N];

void init(){
	REP(i,N){
		upar[i]=i;
	}
}

int fpar(int v){
	if (upar[v]==v){
		return v;
	}
	return upar[v]=fpar(upar[v]);
}

void unite(int s,int t){
	s=fpar(s);
	t=fpar(t);
	upar[t]=s;
}

void dfs(int v,int p){
	for(auto e:tre[v]){
		if (e!=p){
			dfs(e,v);
		}
	}
	bsm[v]+=B[v];
	for(auto e:tre[v]){
		if (e!=p){
			bsm[v]+=bsm[e];
		}
	}
	dp[v]=C[v]+bsm[v];
	for(auto e:tre[v]){
		if (e!=p){
			dp[v]=min(dp[v],max(dp[e],C[v])+bsm[v]-bsm[e]);
		}
	}
}

int main(){
	cin>>N>>M;
	REP(i,N){
		cin>>A[i]>>B[i];
		C[i]=max(A[i]-B[i],0ll);
	}
	REP(i,M){
		int u,v;
		cin>>u>>v;
		u--;
		v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	init();
	REP(i,N){
		tpar[i]=i;
	}
	{
		vector<bool> flg;
		vector<int> vec;
		REP(i,N){
			vec.push_back(i);
			flg.push_back(false);
		}
		sort(vec.begin(),vec.end(),[](int a,int b){
			return C[a]<C[b];
		});
		for(auto v:vec){
			for(auto e:G[v]){
				if (C[v]>C[e] || (C[v]==C[e] && flg[e])){
					tpar[fpar(e)]=v;
					upar[fpar(e)]=v;
				}
			}
			flg[v]=true;
		}
	}
	int rot;
	REP(v,N){
		if (tpar[v]==v){
			rot=v;
		}else{
			tre[v].push_back(tpar[v]);
			tre[tpar[v]].push_back(v);
		}
	}
	dfs(rot,-1);
	cout<<dp[rot]<<endl;
	return 0;
}