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<cstdio>
#include<algorithm>
#include<queue>
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
typedef long long ll;
const int N=100100;
const ll INF=1e18;
int n,m,u,v,o[N],p[N],fa[N],fd[N];
ll ans,a[N],b[N],f[N],sz[N];
vector<int>ad[N];
void add(int u,int v){ad[u].push_back(v);}
int fi(int x){return x==fd[x]?x:fd[x]=fi(fd[x]);}
bool cmp(int x,int y){return a[x]<a[y];}
int main(){
	//freopen("1.in","r",stdin);
	scanf("%d%d",&n,&m);
	FOR(i,1,n) scanf("%lld%lld",&a[i],&b[i]),a[i]-=b[i],o[i]=i;
	FOR(i,1,m) scanf("%d%d",&u,&v),add(u,v),add(v,u);
	sort(o+1,o+n+1,cmp);
	FOR(i,1,n) p[o[i]]=i,fd[i]=i;
	FOR(i,1,n){
		u=o[i];sz[u]=b[u];
		for(int x:ad[u])if(p[v=fi(x)]<i)
			fa[v]=fd[v]=u,sz[u]+=sz[v];
	}
	ans=INF;
	ROF(i,n,1){
		u=o[i];
		f[u]=max(f[fa[u]],a[fa[u]]-sz[u]);
		ans=min(ans,max(a[u],f[u]));
	}
	cout<<ans+sz[o[n]]<<'\n';
}