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
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<vector>
#define ll long long
#define inf 20021225
#define N 100010
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
vector<int> E[N];
struct edge{int to,lt;}e[N]; int in[N],cnt;
void add(int x,int y){e[++cnt].to=y, e[cnt].lt=in[x], in[x]=cnt;}
int a[N],b[N],id[N],fa[N],rk[N],n,m; ll sum[N],f[N];
bool cmp(int x,int y){return a[x]<a[y];}
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
void dfs(int x,int fr)
{
	sum[x]=b[x];
	for(int i=in[x];i;i=e[i].lt)
		dfs(e[i].to,x),sum[x]+=sum[e[i].to]; 
}
void solve(int x,int fr)
{
	if(!in[x])	return void(f[x]=a[x]+b[x]); f[x]=1e18;
	for(int i=in[x];i;i=e[i].lt)
	{
		int y=e[i].to; solve(y,x);
		f[x]=min(f[x],sum[x]-sum[y]+max(1ll*a[x],f[y]));
	}
}
int main()
{
	n=read(),m=read(); int x,y;
	for(int i=1;i<=n;i++)	a[i]=read(),b[i]=read(),a[i]=max(a[i]-b[i],0),id[i]=i;
	for(int i=1;i<=m;i++)	x=read(),y=read(),E[x].push_back(y),E[y].push_back(x);
	sort(id+1,id+n+1,cmp);
	for(int i=1;i<=n;i++)	rk[id[i]]=i,fa[i]=i;
	for(int i=1;i<=n;i++)
	{
		x=id[i];
		for(int j=0;j<E[x].size();j++)
		{
			y=E[x][j]; y=find(y);
			if(rk[y]<rk[x]) add(x,y),fa[y]=x;
		}
	}
	dfs(id[n],0); solve(id[n],0); printf("%lld\n",f[id[n]]);
	return 0;
}