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 MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

ll a[100005],b[100005];

// union-find-simple-ver

struct uftree{
	int par[200005];
	int rank[200005];
	ll dp[200005];
	ll bsum[200005];
	uftree(){
	}
	void init(int n){
		for(int i=0;i<n;i++){
			par[i]=i;
			rank[i]=0;
		}
	}
	int find(int x){
		if(par[x]==x)return x;
		return par[x]=find(par[x]);
	}

	void unite(int x,int y){
		x=find(x);
		y=find(y);
		if(x==y)return;
		if(rank[x]<rank[y]){
			par[x]=y;
			bsum[y]+=bsum[x];
		}else{
			if(rank[x]==rank[y])rank[x]++;
			par[y]=x;
			bsum[x]+=bsum[y];
		}
	}

	bool same(int x,int y){
		return find(x)==find(y);
	}
};


uftree uf;
int n,m;

vector<int> G[100005];
bool flag[200005];
ll maxi=0;
vector<P> vi;

int main(void){
	scanf("%d%d",&n,&m);
	ll all=0;
	uf.init(n);
	for(int i=0;i<n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		all+=b[i];
		vi.push_back(P(max(0LL,a[i]-b[i]),i));
		uf.bsum[i]=b[i];
	}
	
	for(int i=0;i<m;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		u--;
		v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	sort(vi.begin(),vi.end());
	for(int i=0;i<vi.size();i++){
		ll mini=vi[i].first;
		int v=vi[i].second;
		for(int j=0;j<G[v].size();j++){
			if(!flag[G[v][j]])continue;
			int nv=uf.find(G[v][j]);
			mini=min(mini,max(uf.dp[nv],(ll)vi[i].first-uf.bsum[nv]));
		}
		flag[v]=true;
		for(int j=0;j<G[v].size();j++){
			if(flag[G[v][j]]){
				uf.unite(v,G[v][j]);
			}
		}
		uf.dp[uf.find(v)]=mini;
	}
	printf("%lld\n",uf.dp[uf.find(0)]+all);
	return 0;
}