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;
inline int read(){
    register int x=0;
    register char ch=getchar();
    while(ch<'0'||ch>'9')   ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+(ch^48),ch=getchar();
    return x;
}
const int N=2e5+5;
int n,m,now,a[N],b[N],v[N],fa[N],id[N],vi[N],nex[N],head[N];
void addedges(int x,int y){
    nex[++now]=head[x];
    head[x]=now,v[now]=y;
}
vector<int>d[N];
long long f[N],s[N];
bool com(int x,int y){
    return a[x]<a[y];
}
int find(int x){
    return fa[x]?fa[x]=find(fa[x]):x;
}
void dfs(int x){
    int si=d[x].size();
    for(int i=0;i<si;i++){
        int v=d[x][i];
        dfs(v),s[x]+=s[v];
    }
    s[x]+=b[x],f[x]=s[x]+a[x];
    for(int i=0;i<si;i++){
        int v=d[x][i];
        f[x]=min(f[x],s[x]-s[v]+max(f[v],1LL*a[x]));
    }
}
int main(){
    //freopen("1.out","w",stdout);
    //freopen("1.in","r",stdin);
    n=read(),m=read();
    for(int i=1;i<=n;i++)   a[i]=read(),b[i]=read(),a[i]=max(a[i]-b[i],0);
    for(int i=1;i<=m;i++){
        int x=read(),y=read();
        addedges(x,y),addedges(y,x);
    }
    for(int i=1;i<=n;i++)   id[i]=i;
    sort(id+1,id+n+1,com);
    for(int i=1;i<=n;i++){
        int x=id[i];    vi[x]=1;
        for(int j=head[x];j;j=nex[j])
            if(vi[v[j]]){
                int y=find(v[j]);
                if(y==x)    continue;
                d[x].push_back(y),fa[y]=x;
            }
    }
    dfs(id[n]);
    printf("%lld\n",f[id[n]]);
    return 0;
}