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;
typedef long long ll;
const ll MAX=100020;
const ll INF=1ll<<62;
ll N,M,A[MAX],B[MAX],Q[MAX],par[MAX],W[MAX],Btot[MAX],chk[MAX];
vector<ll> conn[MAX],rec;
bool cmp(ll x,ll y){
    return A[x]-B[x]<A[y]-B[y];
}
ll fnd(ll x){
    if(par[x]==x) return x;
    return par[x]=fnd(par[x]);
}
ll uni(ll x,ll y){
    x=fnd(x); y=fnd(y);
    if(x==y) return 0;
    par[x]=y; return 1;
}
int main(){
    scanf("%lld%lld",&N,&M);
    for(ll i=1;i<=N;i++) scanf("%lld%lld",&A[i],&B[i]);
    for(ll i=0;i<M;i++){
        ll t1,t2;
        scanf("%lld%lld",&t1,&t2);
        conn[t1].push_back(t2);
        conn[t2].push_back(t1);
    }
    for(ll i=1;i<=N;i++) Q[i]=i,par[i]=i;
    sort(Q+1,Q+1+N,cmp);
    for(ll i=1;i<=N;i++){
        ll cnt=0,idx;
        chk[Q[i]]=1;
        W[Q[i]]=INF;
        for(ll j=0;j<conn[Q[i]].size();j++) if(chk[conn[Q[i]][j]]) {
            idx=fnd(conn[Q[i]][j]);
            if(idx!=Q[i]) rec.push_back(idx), par[idx]=Q[i];
        }
        for(ll j=0;j<rec.size();j++) Btot[Q[i]]+=Btot[rec[j]];
        if(rec.size()==0) W[Q[i]]=max(A[Q[i]],B[Q[i]]);
        else {for(ll j=0;j<rec.size();j++) W[Q[i]]=min(W[Q[i]],Btot[Q[i]]-Btot[rec[j]]+max(A[Q[i]],B[Q[i]]+W[rec[j]]));}
        Btot[Q[i]]+=B[Q[i]];
        rec.clear();
    }
    printf("%lld\n",W[Q[N]]);
}
