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 int long long
const int N=200005;
int flag[N],f[N],c[N],dp[N],x,y,n,m,id[N],rk[N],ne[N],fi[N],zz[N],tot,size[N],a[N],b[N];
vector<int> G[N];
int find(int x){
	if (x==f[x])return x;
	return f[x]=find(f[x]);
}
int cmp(int x,int y){
	return c[x]<c[y];
}
void jb(int x,int y){
	ne[++tot]=fi[x];
	fi[x]=tot;
	zz[tot]=y;
}
void dfs1(int x){
	size[x]=b[x];
	for (int i=0;i<G[x].size();i++)dfs1(G[x][i]),size[x]+=size[G[x][i]];
}
void dfs2(int x){
	if (!G[x].size()){
		dp[x]=b[x]+c[x];
		return;
	}
	dp[x]=1e18;
	for (int i=0;i<G[x].size();i++)dfs2(G[x][i]),dp[x]=min(dp[x],max(c[x],dp[G[x][i]])+size[x]-size[G[x][i]]);
}
signed main(){
	scanf("%lld%lld",&n,&m);
	for (int i=1;i<=n;i++)scanf("%lld%lld",&a[i],&b[i]),c[i]=max(a[i]-b[i],0ll);
	for (int i=1;i<=n;i++)f[i]=i,id[i]=i;
	while (m--){
		scanf("%lld%lld",&x,&y);
		jb(x,y);jb(y,x);
	}
	sort(id+1,id+n+1,cmp);
	for (int i=1;i<=n;i++)rk[id[i]]=i;
	for (int i=1;i<=n;i++)
		for (int j=fi[id[i]];j;j=ne[j]){
			x=find(id[i]),y=find(zz[j]);
			if (id[i]!=y&&rk[x]>rk[y])G[id[i]].push_back(y),f[y]=id[i],flag[y]=1;
		}
	int root=0;
	for (int i=1;i<=n;i++)
		if (!flag[i])root=i;
	dfs1(root);
	dfs2(root);
	printf("%lld\n",dp[root]);
}
