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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=1000005;
struct Edge{int u,v;int w;
inline bool operator <(const Edge &b)const {return w<b.w;}
}e[N];
ll f[N],sum[N];
int fa[N];
int a[N],b[N];
int find(int x){return fa[x]=(fa[x]==x?x:find(fa[x]));}
int main()
{
	//freopen("input.txt","r",stdin);
	ios::sync_with_stdio(false);
	cin.tie(0);
	int n,m,tmp=0;
	cin >>n>>m;
	for(int i=1;i<=n;i++)cin >>a[i]>>b[i],fa[i]=i,sum[i]=b[i],f[i]=max(a[i],b[i]);
	for(int i=1;i<=m;i++)cin >>e[i].u>>e[i].v,e[i].w=max(max(a[e[i].u]-b[e[i].u],0),max(a[e[i].v]-b[e[i].v],0));
	sort(e+1,e+1+m);
	for(int i=1;i<=m;i++)if((e[i].u=find(e[i].u))!=(e[i].v=find(e[i].v)))
	{
		f[e[i].u]=min(max(f[e[i].u],(ll)e[i].w)+sum[e[i].v],max(f[e[i].v],(ll)e[i].w)+sum[e[i].u]);
		fa[e[i].v]=e[i].u;
		sum[e[i].u]+=sum[e[i].v];
	}
	cout <<f[find(1)]<<endl;
	//cout <<tmp<<n;
	return 0;
}