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<stdio.h>
#include<algorithm>
using namespace std;
typedef long long ll;
struct node{
    int x,v;
    node(int a=0,int b=0){x=a;v=b;}
}p[100010];
int h[100010],nex[200010],to[200010],a[100010],b[100010],fa[100010],M;
ll s[100010],f[100010];
bool v[100010];
void add(int a,int b){
    M++;
    to[M]=b;
    nex[M]=h[a];
    h[a]=M;
}
bool operator<(node a,node b){return a.v<b.v;}
int get(int x){return x==fa[x]?x:(fa[x]=get(fa[x]));}
int main(){
    int n,m,i,j,x,y;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++){
        scanf("%d%d",a+i,b+i);
        a[i]=max(a[i]-b[i],0);
        p[i]=node(i,a[i]);
    }
    while(m--){
        scanf("%d%d",&x,&y);
        add(x,y);
        add(y,x);
    }
    sort(p+1,p+n+1);
    for(i=1;i<=n;i++){
        fa[i]=i;
        s[i]=b[i];
        f[i]=a[i];
    }
    for(i=1;i<=n;i++){
        v[p[i].x]=1;
        for(j=h[p[i].x];j;j=nex[j]){
            if(v[to[j]]){
                x=get(p[i].x);
                y=get(to[j]);
                if(x==y)continue;
                fa[y]=x;
                s[x]+=s[y];
                f[x]=min(f[x],max(f[y],a[x]-s[y]));
            }
        }
    }
    x=get(1);
    printf("%lld",f[x]+s[x]);
}