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 MN 100000
#define ll long long
using namespace std;
inline int read()
{
	int x=0;char ch=getchar();
	while(ch<'0'||ch>'9')ch=getchar();
	while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
	return x;
}
ll f[MN+5],g[MN+5];
int n,m,a[MN+5],b[MN+5],c[MN+5],s[MN+5],rk[MN+5],used[MN+5],head[MN+5],cnt;
struct edge{int to,next;}e[MN*2+5];
inline void ins(int f,int t)
{
	e[++cnt]=(edge){t,head[f]};head[f]=cnt;
	e[++cnt]=(edge){f,head[t]};head[t]=cnt;	
}
inline int getfa(int x){return s[x]?s[x]=getfa(s[x]):x;}
bool cmp(int x,int y){return c[x]<c[y];}
int main()
{
	n=read();m=read();
	for(int i=1;i<=n;++i) a[i]=read(),b[i]=read(),c[i]=max(a[i]-b[i],0),rk[i]=i;
	sort(rk+1,rk+n+1,cmp);
	for(int i=1;i<=m;++i) ins(read(),read());
	for(int i=1;i<=n;++i)
	{
		int x=rk[i];used[x]=1;ll mn=c[x];g[x]=b[x];
		for(int j=head[x];j;j=e[j].next) if(used[e[j].to])
		{
			int y=getfa(e[j].to);
			if(y==x) continue;
			s[y]=x;g[x]+=g[y];
			mn=min(mn,max((ll)c[x],f[y])-g[y]);
		}
		f[x]=g[x]+mn;
	}
	printf("%lld",f[rk[n]]);
	return 0;
}

