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;

const int N=100010;
int n,m;
struct edge{
	int y,next;
}s[N<<1];
vector<int> out[N];
int first[N],len=0,a[N],b[N],id[N],rk[N],fa[N];
long long sum[N],dp[N];

bool cmp(int x,int y){return a[x]<a[y];}
int findpa(int x){return fa[x]!=x?fa[x]=findpa(fa[x]):x;}
void ins(int x,int y){s[++len]=(edge){y,first[x]};first[x]=len;}

void dfs(int x,int fa){
	sum[x]=b[x];
	for(int i=first[x];i!=0;i=s[i].next) if(s[i].y!=fa) dfs(s[i].y,x),sum[x]+=sum[s[i].y];
}

void get_ans(int x,int fa){
	if(first[x]==0) dp[x]=a[x]+b[x];
	else dp[x]=1e18;
	for(int i=first[x];i!=0;i=s[i].next) if(s[i].y!=fa){
		get_ans(s[i].y,x);
		dp[x]=min(dp[x],sum[x]-sum[s[i].y]+max(dp[s[i].y],1ll*a[x]));
	}
}

int main(){
	scanf("%d %d",&n,&m);
	int x,y;
	for(int i=1;i<=n;i++) scanf("%d %d",&a[i],&b[i]),a[i]=max(a[i]-b[i],0),id[i]=i;
	for(int i=1;i<=m;i++) scanf("%d %d",&x,&y),out[x].push_back(y),out[y].push_back(x);
	sort(id+1,id+1+n,cmp);
	for(int i=1;i<=n;i++) rk[id[i]]=i,fa[i]=i;
	for(int i=1;i<=n;i++){
		x=id[i];
		for(int j=0;j<out[x].size();j++){
			y=findpa(out[x][j]);
			if(rk[y]<rk[x]) ins(x,y),fa[y]=x;
		}
	}
	dfs(id[n],0);
	get_ans(id[n],0);
	printf("%lld\n",dp[id[n]]);
}