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>
    #define LL long long
    #define db long double
     
    using namespace std;
    const int N=3e5+10;
    int rd()
    {
    	int x=0,w=1;char ch=0;
    	while(ch<'0'||ch>'9'){if(ch=='-') w=-1;ch=getchar();}
    	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}
    	return x*w;
    }
    int to[N<<1],nt[N<<1],hd[N],tot=1;
    void adde(int x,int y)
    {
    	++tot,to[tot]=y,nt[tot]=hd[x],hd[x]=tot;
    	++tot,to[tot]=x,nt[tot]=hd[y],hd[y]=tot;
    }
    int n,m,sq[N],ff[N],st[N],tp;
    LL a[N],b[N],f[N],g[N];
    bool vs[N];
    bool cmp(int aa,int bb){return a[aa]-b[aa]<a[bb]-b[bb];}
    int findf(int x){return ff[x]==x?x:ff[x]=findf(ff[x]);}
     
    int main()
    {
    	n=rd(),m=rd();
    	for(int i=1;i<=n;++i) a[i]=rd(),b[i]=rd(),sq[i]=ff[i]=i;
    	sort(sq+1,sq+n+1,cmp);
    	for(int i=1;i<=m;++i) adde(rd(),rd());
    	for(int i=1;i<=n;++i)
    	{
    		int x=sq[i];
    		g[x]=b[x],vs[x]=1;
    		for(int j=hd[x];j;j=nt[j])
    		{
    			int y=findf(to[j]);
    			if(!vs[y]||y==x) continue;
    			ff[y]=x,g[x]+=g[y],st[++tp]=y;
    		}
    		if(!tp) f[x]=max(a[x],b[x]);
    		else
    		{
    			f[x]=1ll<<50;
    			while(tp)
    			{
    				int y=st[tp--];
    				f[x]=min(f[x],-g[y]+max(f[y],max(a[x]-b[x],0ll)));
    			}
    			f[x]+=g[x];
    		}
    	}
    	printf("%lld\n",f[sq[n]]);
    	return 0;
    }