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;
#define ll long long
const int maxn=1e5+7;
bool vis[maxn];
int a[maxn],b[maxn],p[maxn],fa[maxn];
ll c[maxn];
ll dp[maxn],sum[maxn];
struct Edge{
	int nxt,to;
}G[maxn<<1],T[maxn<<1];
int bG=0,hG[maxn],bT=0,hT[maxn<<1];
void addG(int x,int y){
	G[bG].nxt=hG[x];G[bG].to=y;hG[x]=bG++;
	G[bG].nxt=hG[y];G[bG].to=x;hG[y]=bG++;
}
void addT(int x,int y){
	T[bT].nxt=hT[x];T[bT].to=y;hT[x]=bT++;
	T[bT].nxt=hT[y];T[bT].to=x;hT[y]=bT++;
}
bool cmp(int x,int y){
	return c[x]<c[y];
}
void dfs(int v,int p){
	sum[v]=b[v];
	for (int i=hT[v];i!=-1;i=T[i].nxt){
		int u=T[i].to;
		if (u==p) continue;
		dfs(u,v);
		sum[v]+=sum[u];
	}
	dp[v]=sum[v]+c[v];
	for (int i=hT[v];i!=-1;i=T[i].nxt){
		int u=T[i].to;
		if (u==p) continue;
		dp[v]=min(dp[v],sum[v]-sum[u]+max(dp[u],c[v]));
	}
}
int Get(int x){
	if (fa[x]==x) return x;
	fa[x]=Get(fa[x]);return fa[x];
}
int main(){
	memset(hG,-1,sizeof(hG));
	memset(hT,-1,sizeof(hT));
	int n,m;scanf("%d%d",&n,&m);
	for (int i=0;i<n;i++){
		scanf("%d%d",a+i,b+i);
		c[i]=max(a[i]-b[i],0);
		p[i]=i;fa[i]=i;
	}
	for (int i=0;i<m;i++){
		int u,v;scanf("%d%d",&u,&v);u--;v--;addG(u,v);
	}
	sort(p,p+n,cmp);
	for (int i=0;i<n;i++){
		int v=p[i];
		vis[v]=true;
		for (int e=hG[v];e!=-1;e=G[e].nxt){
			int u=G[e].to;
			int fu=Get(u);
			if (vis[u] && v!=fu){
				fa[fu]=v;
				addT(fu,v);
			}
		}
	}
	dfs(p[n-1],-1);
	printf("%lld\n",dp[p[n-1]]);
	return 0;
}