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<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
#define chmin(a,b) a=min(a,b);
#define chmax(a,b) a=max(a,b);
#define N 100010
vector<ll> e[N];
ll n,m,a[N],b[N];

namespace uf{
  ll p[N];
  void init(){
      for(int i=0;i<N;i++)p[i]=i;
  }
  ll par(ll x){return p[x]=(x==p[x]?x:par(p[x]));}
  void unite(ll x,ll y){
      x=par(x),y=par(y);
      if(x!=y){
	  p[x]=y;
      }
  }
  bool same(ll x,ll y){return par(x)==par(y);}
};

ll f[N],g[N];
P dat[N]; bool vis[N];
int main(){
  cin>>n>>m;
  for(int i=1;i<=n;i++){
      cin>>a[i]>>b[i];
      dat[i-1]=make_pair(a[i]-b[i],i);
  }
  sort(dat,dat+n);
  
  for(int i=0;i<m;i++){
      ll x,y;cin>>x>>y;
      e[x].push_back(y);
      e[y].push_back(x);
  }
  
  for(int i=1;i<=n;i++)vis[i]=0,g[i]=b[i];
  uf::init();
  for(int i=0;i<n;i++){
      int y=dat[i].second; vis[y]=1;
      f[y]=1e17; //cout<<"#"<<y<<endl;
      vector<ll> vist;
      for(auto x:e[y])if(vis[x]==1&&uf::same(x,y)==0){
	  x=uf::par(x); vist.push_back(x);
	  g[y]+=g[x];
	  uf::unite(x,y);
      }
      if(g[y]==b[y]){f[y]=max(a[y],b[y]); continue;}
      for(auto x:vist){
	  chmin(f[y],max(a[y],b[y]+f[x])+g[y]-g[x]-b[y]);
      }
  }
  cout<<f[dat[n-1].second]<<endl;
  //for(int i=1;i<=n;i++)cout<<f[i]<<" "<<g[i]<<endl;
    return 0;
}

