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 rep(i,s,t) for(int i=s;i<=(t);++i) 
using namespace std; 

const int N=1e5+50; 

bool vis[N]; 
int n,m,a[N],b[N],p[N],id[N]; 
long long sum[N],f[N]; 
vector<int> g[N];

int find(int x){ return p[x]==x?x:p[x]=find(p[x]); }

int main(int argc,char *argv[]){
	scanf("%d%d",&n,&m); 
	rep(i,1,n){
		scanf("%d%d",&a[i],&b[i]); 
		a[i]=max(0,a[i]-b[i]); 
		sum[i]=b[i];
	}
	rep(i,1,m){
		int u,v; 
		scanf("%d%d",&u,&v); 
		g[u].push_back(v);
		g[v].push_back(u);
	}
	rep(i,1,n)p[i]=i,id[i]=i;
	sort(id+1,id+n+1,[](int x,int y){ return a[x]<a[y]; }); 
	rep(k,1,n){
		int v=id[k]; 
		vis[v]=true;
		vector<int> son;
		for(int d:g[v])if(vis[d]&&find(d)!=find(v)){
			sum[v]+=sum[find(d)];
			son.push_back(find(d)); 
			p[find(d)]=find(v); 
		}
		f[v]=sum[v]+a[v];
		
		for(int d:son){
			f[v]=min(f[v],sum[v]-sum[d]+max(1ll*a[v],f[d])); 
		}
	}
	printf("%lld\n",f[id[n]]);
	return 0; 
}
