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 fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
#define ll long long
using namespace std;
const int N=1e5+5;
int n,m,x,y,a[N],b[N],w[N],f[N];
vector<int>e[N];
ll s[N],g[N];
bool ex[N];
bool cmp(int x,int y){return a[x]<a[y];}
int F(int x){return x==f[x]?x:f[x]=F(f[x]);}
int main(){
	ios::sync_with_stdio(false);
	cin>>n>>m;
	fo(i,1,n)cin>>a[i]>>b[i],a[i]=max(a[i]-b[i],0),w[i]=f[i]=i;
	fo(i,1,m)cin>>x>>y,e[x].push_back(y),e[y].push_back(x);
	sort(w+1,w+n+1,cmp);
	fo(i,1,n){
		int x=w[i],leaf=1;ex[x]=1;
		s[x]=b[x];
		g[x]=1e18;
		for(int j:e[x])if(ex[j]&&F(j)!=x){
			int y=f[j];
			f[y]=x;
			s[x]+=s[y];
			g[x]=min(g[x],max(g[y],-s[y]+a[x]));
			leaf=0;
		}
		if(leaf)g[x]=a[x];
	}
	cout<<g[w[n]]+s[w[n]];
}