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 <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const int N=110000;
int father[N];
long long sum[N],w[N];
int n,m,A[N],B[N],pos[N],in[N];
int findfather(int k1){
	if (father[k1]==k1) return k1; return father[k1]=findfather(father[k1]);
}
vector<int> go[N];
int compare(int k1,int k2){
	return max(0,A[k1]-B[k1])<max(0,A[k2]-B[k2]);
}
int pd[N],sign;
vector<int> C;
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) scanf("%d%d",&A[i],&B[i]);
	for (int i=1;i<=m;i++){
		int k1,k2; scanf("%d%d",&k1,&k2);
		go[k1].push_back(k2);
		go[k2].push_back(k1);
	}
	for (int i=1;i<=n;i++) pos[i]=i,father[i]=i;
	sort(pos+1,pos+n+1,compare);
	for (int i=1;i<=n;i++){
		int k1=pos[i]; sign++; C.clear();
		for (int j=0;j<go[k1].size();j++){
			int k2=go[k1][j];
			if (in[k2]){
				int root=findfather(k2);
				if (pd[root]!=sign){
					pd[root]=sign; C.push_back(root);
				}
			}
		}
		for (int j=0;j<C.size();j++){
			father[C[j]]=k1;
			sum[k1]+=sum[C[j]];
		} in[k1]=1;
		if (C.size()==0){
			w[k1]=max(A[k1],B[k1]); sum[k1]=B[k1]; continue;
		}
		w[k1]=sum[k1]+max(A[k1],B[k1]);
		for (int j=0;j<C.size();j++)
			w[k1]=min(w[k1],sum[k1]-sum[C[j]]+max(1ll*A[k1],w[C[j]]+B[k1]));
		sum[k1]+=B[k1];
		//cout<<k1<<" "<<w[k1]<<" "<<B[k1]<<endl;
	}
	printf("%lld\n",w[findfather(1)]);
}