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>
using namespace std;
const int maxn=1e5+10;
int n,m;
int a[maxn],b[maxn],id[maxn];
struct edge{int v,nxt;}e[maxn<<1];
int head[maxn],ecnt=0,vis[maxn];
void add(int u,int v)
{
	e[++ecnt]=(edge){v,head[u]},head[u]=ecnt;
	e[++ecnt]=(edge){u,head[v]},head[v]=ecnt;
}
bool cmp(const int &x,const int &y){return a[x]<a[y];}
vector<int>G[maxn];
int fa[maxn];
long long sum[maxn],dp[maxn];
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
void dfs(int u,int pre)
{
	
	for(int i=head[u];~i;i=e[i].nxt)
	{
		int v=e[i].v;
		if(v==pre)continue;
		dfs(v,u);
		sum[u]+=sum[v];
	}
	sum[u]+=b[u];
	dp[u]=sum[u]+a[u];
	for(int i=head[u];~i;i=e[i].nxt)
	{
		int v=e[i].v;
		if(v==pre)continue;
		dp[u]=min(dp[u],sum[u]-sum[v]+max(dp[v],a[u]*1LL));
	}
}
int main()
{
	memset(head,-1,sizeof(head));
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),a[i]=max(0,a[i]-b[i]),id[i]=i;
	sort(id+1,id+1+n,cmp);
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<=m;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for(int i=1;i<=n;i++)
	{
		int u=id[i];
		vis[u]=1;
		for(auto v:G[u])
		if(vis[v]&&find(u)!=find(v))
		{
			add(u,find(v));
			fa[find(v)]=u;
		}
	}
	dfs(id[n],0);
	printf("%lld\n",dp[id[n]]);
}