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>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
using namespace std;
const int N=100005;
struct edge{
	int to,next;
}e[N*2];
int head[N],tot,n,m;
int A[N],B[N],C[N],id[N];
int fa[N],vis[N];
ll mx[N],S[N];
bool cmp(int x,int y){
	return C[x]<C[y];
}
int get(int x){
	return x==fa[x]?x:fa[x]=get(fa[x]);
}
void add(int x,int y){
	e[++tot]=(edge){y,head[x]};
	head[x]=tot;
}
int main(){
	scanf("%d%d",&n,&m);
	For(i,1,n) scanf("%d%d",&A[i],&B[i]);
	For(i,1,n) C[i]=max(0,A[i]-B[i]),id[i]=i;
	For(i,1,m){
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	sort(id+1,id+n+1,cmp);
	For(i,1,n) fa[i]=i;
	For(i,1,n){
		int x=id[i]; vis[x]=1;
		mx[x]=C[x]; S[x]=B[x];
		for (int j=head[x];j;j=e[j].next)
			if (vis[e[j].to]&&get(e[j].to)!=get(x)){
				int y=get(e[j].to);
				mx[x]=min(mx[x],mx[y]+max(0ll,C[x]-(S[y]+mx[y])));
				S[x]=S[x]+S[y]; fa[y]=x;
			}
	}
	printf("%lld",S[get(1)]+mx[get(1)]);
}