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;
typedef long long ll;
const int N=100005,inf=0x3f3f3f3f;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,m,a[N],b[N],pos[N],fa[N],h[N];
ll f[N],g[N];
vector<int> e[N];
bool cmp(int x,int y){return a[x]<a[y];}
int find(int x){return (x==fa[x]) ? x : fa[x]=find(fa[x]);}
void merge(int x,int y){
	y=find(y);fa[y]=x;
	g[x]+=g[y];
	f[x]=min(f[x],max(f[y],a[x]-g[y]));
}
int main(){
	n=read();m=read();
	for (int i=1;i<=n;i++){
		a[i]=read();b[i]=read();a[i]=max(a[i]-b[i],0);
		pos[i]=fa[i]=i;f[i]=a[i];g[i]=b[i];
	}
	while (m--){
		int u=read(),v=read();
		e[u].push_back(v);
		e[v].push_back(u);
	}
	sort(pos+1,pos+n+1,cmp);
	for (int i=1;i<=n;i++){
		int x=pos[i];h[x]=1;
		for (auto y : e[x])
		if (h[y]&&(find(y)!=x)) merge(x,y);
	}
	printf("%lld\n",f[find(1)]+g[find(1)]);
	return 0;
}