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 N=500005,inf=1e9;
typedef long long ll;
int read(){
	int x=0,f=1; char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1; ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
	return x*f;
}
struct Edg{
	int nxt,poi;
}e[N<<1];
int fa[N],a[N],b[N],first[N],l=0,c[N];
ll f[N],g[N];
bool vis[N];
bool cmp(int x,int y){
	return a[x]<a[y];
}
void addedge(int u,int v){
	l++;
	e[l].nxt=first[u];
	e[l].poi=v;
	first[u]=l;
}
int find(int x){
	return (fa[x]==x)?x:fa[x]=find(fa[x]);
}
int main(){
	int n=read(),m=read();
	for (int i=1;i<=n;i++){
		a[i]=read(),b[i]=read(); a[i]=max(0,a[i]-b[i]);
	}
	for (int i=1;i<=m;i++){
		int u=read(),v=read();
		addedge(u,v); addedge(v,u);
	}
	for (int i=1;i<=n;i++){
		fa[i]=i; c[i]=i; f[i]=a[i]; g[i]=b[i];
	}
	sort(c+1,c+1+n,cmp);
	for (int i=1;i<=n;i++){
		int u=c[i]; vis[u]=1;
		for (int p=first[u];p;p=e[p].nxt){
			int v=e[p].poi;
			if (!vis[v]) continue;
			int x=find(u),y=find(v);
			if (x==y) continue;
			g[x]+=g[y]; fa[y]=x;
			f[x]=min(f[x],f[y]+max(0ll,a[u]-f[y]-g[y]));
		}
	}
	printf("%lld\n",f[c[n]]+g[c[n]]);
	return 0;
}