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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=100005;
int n,m,a[N],b[N],head[N],adj[N*2],nxt[N*2],c[N],p[N],l,Head[N],Adj[N*2],Nxt[N*2],id[N],i,j,k,fa[N];
ll sum[N],f[N];
int find(int x)
{
	return !fa[x]?x:fa[x]=find(fa[x]);
}
bool cmp(int a,int b)
{
	return c[a]<c[b];
}
void Addedge(int u,int v)
{
	Adj[++l]=v;
	Nxt[l]=Head[u];
	Head[u]=l;
}
void dfs(int x)
{
	int j;
	sum[x]=b[x];
	for(j=Head[x];j;j=Nxt[j])
	{
		dfs(Adj[j]);
		sum[x]+=sum[Adj[j]];
	}
	f[x]=sum[x]+c[x];
	for(j=Head[x];j;j=Nxt[j])
		f[x]=min(f[x],sum[x]-sum[Adj[j]]+max(1ll*c[x],f[Adj[j]]));
}
int main()
{
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;++i)
	{
		scanf("%d%d",a+i,b+i);
		c[i]=max(0,a[i]-b[i]);
	}
	for(i=1;i<=m;++i)
	{
		scanf("%d%d",&j,&k);
		adj[i*2-1]=k;
		nxt[i*2-1]=head[j];
		head[j]=i*2-1;
		adj[i*2]=j;
		nxt[i*2]=head[k];
		head[k]=i*2;
	}
	for(i=1;i<=n;++i)
		p[i]=i;
	sort(p+1,p+1+n,cmp);
	for(i=1;i<=n;++i)
		id[p[i]]=i;
	for(i=1;i<=n;++i)
		for(j=head[p[i]];j;j=nxt[j])
			if(id[adj[j]]<i&&find(adj[j])!=find(p[i]))
			{
				Addedge(p[i],find(adj[j]));
				fa[find(adj[j])]=p[i];
			}
	dfs(p[n]);
	printf("%lld",f[p[n]]);
	return 0;
}
