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<stdio.h>
#include<algorithm>
using namespace std;
int head2[131072],last2[131072],to2[131072],cnt2=0;
void add2(int u,int v)
{
	cnt2++;
	last2[cnt2]=head2[u];
	head2[u]=cnt2;
	to2[cnt2]=v;
}
int a[131072],b[131072];
long long s[131072],ans[131072];
long long dfs(int u)
{
	s[u]=a[u]-b[u];
	for(int i=head2[u];i;i=last2[i])
	{
		int v=to2[i];
		dfs(v);
		s[u]+=s[v];
	}
	ans[u]=s[u]+b[u];
	for(int i=head2[u];i;i=last2[i])
	{
		int v=to2[i];
		long long a2=s[u]-s[v];
		if(ans[v]>b[u])
		{
			a2+=ans[v];
		}
		else
		{
			a2+=b[u];
		}
		if(a2<ans[u])
		{
			ans[u]=a2;
		}
	}
	return ans[u];
}
int head[131072],last[262144],to[262144],cnt=0;
void add(int u,int v)
{
	cnt++;
	last[cnt]=head[u];
	head[u]=cnt;
	to[cnt]=v;
}
struct node{
	int u;
	int a;
	int b;
	bool operator<(node c) const
	{
		return b<c.b;
	}
};
node nn[131072];
bool vis[131072];
int fa[131072];
int findroot(int x)
{
	if(fa[x]==x)
	{
		return x;
	}
	return fa[x]=findroot(fa[x]);
}
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
	{
		nn[i].u=i+1;
		int a,b;
		scanf("%d%d",&a,&b);
		if(a<b)
		{
			a=b;
		}
		nn[i].a=a;
		nn[i].b=a-b;
	}
	sort(nn,nn+n);
	while(m--)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
	}
	for(int i=1;i<=n;i++)
	{
		fa[i]=i;
		vis[i]=false;
	}
	for(int t=0;t<n;t++)
	{
		int u=nn[t].u;
		vis[u]=true;
		a[u]=nn[t].a;
		b[u]=nn[t].b;
		for(int i=head[u];i;i=last[i])
		{
			int v=to[i];
			if(!vis[v])
			{
				continue;
			}
			v=findroot(v);
			if(v!=u)
			{
				add2(u,v);
				fa[v]=u;
			}
		}
	}
	int root=nn[n-1].u;
	printf("%lld\n",dfs(root));
	return 0;
}