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<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<vector>
#define ll long long
using namespace std;

const int N=100005;

int n,m,x,y,cnt;
int a[N],b[N],fa[N],id[N],son[N],vis[N];
ll dp[N],sum[N];
vector<int> lev[N];

bool cmp(int x,int y){
	return a[x]<a[y];
}

int find(int k){
	if (fa[k]!=k) fa[k]=find(fa[k]);
	return fa[k];
}

int main(){
	int i,j;
	scanf("%d%d",&n,&m);
	for (i=1;i<=n;i++){
		scanf("%d%d",&a[i],&b[i]);
		a[i]=max(a[i]-b[i],0);
		id[i]=i;
	}
	for (i=1;i<=m;i++){
		scanf("%d%d",&x,&y);
		lev[x].push_back(y);
		lev[y].push_back(x);
	}
	sort(id+1,id+1+n,cmp);
	for (i=1;i<=n;i++) fa[i]=i;
	for (i=1;i<=n;i++){
		cnt=0; x=id[i]; vis[x]=1; sum[x]=1ll*b[x];
		for (j=0;j<lev[x].size();j++){
			y=lev[x][j];
			if (!vis[y]||find(x)==find(y)) continue;
			son[++cnt]=find(y);
			sum[x]+=1ll*sum[find(y)];
			fa[find(y)]=find(x);
		}
		dp[x]=sum[x]+1ll*a[x];
		for (j=1;j<=cnt;j++) dp[x]=min(dp[x],sum[x]-sum[son[j]]+max(1ll*a[x],dp[son[j]]));
	}
	printf("%lld\n",dp[id[n]]);
	return 0;
}