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;

#define ll long long
int n, m, id[101000]; ll a[101000], b[101000], f[101000], sb[101000];
vector<int>G[101000];
int fa[101000], fl[101000];
int GF(int x){
    return fa[x]==x? x: fa[x]=GF(fa[x]);
}
int main(){
    cin>>n>>m;
    for (int i=1;i<=n;++i){
        scanf("%lld%lld",&a[i],&b[i]);
        a[i]=max(a[i],b[i])-b[i];
    }
    for (int x,y;m--;){
        scanf("%d%d",&x,&y);
        G[x].push_back(y);
        G[y].push_back(x);
    }
    for (int i=1;i<=n;++i){
        fa[i]=i; id[i]=i;
    }
    sort(id+1,id+n+1,[](int i,int j){return a[i]<a[j];});
    for (int i=1;i<=n;++i){
        int x=id[i]; fl[x]=1; sb[x]=b[x];
        vector<int>ch;
        for (auto y:G[x]){
            if (!fl[y=GF(y)]) continue;
            if (y==x) continue;
            ch.push_back(y); fa[y]=x;
        }
        for (auto y:ch) sb[x]+=sb[y];
        f[x]=sb[x]+a[x];
        for (auto y:ch) f[x]=min(f[x],sb[x]-sb[y]+max(a[x],f[y]));
        //printf(" %d %lld %lld\n",x,sb[x],f[x]);
    }
    cout<<*max_element(f+1,f+n+1)<<endl;
}