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<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;
typedef long long LL;
const int _=1e2;
const int maxn=1e5+_;
const int maxm=1e5+_;

int A[maxn],B[maxn],id[maxn];
bool cmp(int n1,int n2){return A[n1]<A[n2];}
vector<int>vec[maxn];
struct node
{
	int x,y,next;
}a[maxm];int len,last[maxn];
void ins(int x,int y)
{
	len++;
	a[len].x=x;a[len].y=y;
	a[len].next=last[x];last[x]=len;
}
int F[maxn];
LL f[maxn],sum[maxn];//进入这棵树搞完最少要多少速度，这棵树里面的总消耗速度
void dfs(int x)
{
	if(last[x]==0){f[x]=A[x]+B[x];sum[x]=B[x];return ;}
	sum[x]=B[x];
	for(int k=last[x];k;k=a[k].next)
	{
		int y=a[k].y;
		dfs(y);
		sum[x]+=sum[y];
	}
	f[x]=(1LL<<62);
	for(int k=last[x];k;k=a[k].next)
	{
		int y=a[k].y;
		f[x]=min(f[x],max(f[y],(LL)A[x])+sum[x]-sum[y]);
	}
}

int fa[maxn];
int findfa(int x)
{
	if(x==fa[x])return x;
	fa[x]=findfa(fa[x]);return fa[x];
}
int main()
{
	int n,m,x,y;
	scanf("%d%d",&n,&m); len=1;
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&A[i],&B[i]);
		A[i]=max(A[i]-B[i],0);
		id[i]=i;
	}
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d",&x,&y);
		vec[x].push_back(y);
		vec[y].push_back(x);
	}
		
	sort(id+1,id+n+1,cmp);
	memset(F,-1,sizeof(F));
	for(int t=1;t<=n;t++)
	{
		int i=id[t]; fa[i]=i;
		for(int k=0;k<vec[i].size();k++)
		{
			int j=vec[i][k],fj=findfa(j);
			if(F[fj]==0)F[fj]=i,fa[fj]=i;
		}
		F[i]=0;
	}
	int rt;
	for(int i=1;i<=n;i++)
		if(F[i]==0)rt=i;
		else ins(F[i],i);
		
	dfs(rt);
	printf("%lld\n",f[rt]);
		
	return 0;
}