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=100100;
int a[N],b[N],n,m,vis[N],fa[N],c[N];
typedef long long ll;
vector<int>g[N];
ll sum[N],mn[N];
bool cmp(const int&x,const int&y){
	if(a[x]!=a[y])return a[x]<a[y];
	return b[x]>b[y];
}
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=1;i<=n;++i){cin>>a[i]>>b[i];c[i]=i;}
	sort(c+1,c+n+1,cmp);
	for(int i=1;i<=m;++i){
		int x,y;cin>>x>>y;
		g[x].push_back(y);g[y].push_back(x);
	}
	for(int i=1;i<=n;++i)fa[i]=i;//,sort(g[i].begin(),g[i].end(),cmp);
	for(int i=1;i<=n;++i){
		int u=c[i];sum[u]=b[u];mn[u]=max(0,a[u]-b[u]);
		vis[u]=1;
		for(int j:g[u])if(vis[j]){
			int v=find(j);if(u==v)continue;
			sum[u]+=sum[v];mn[u]=min(mn[u],max(a[u]-sum[v]-b[u],mn[v]));
			fa[v]=u;
		}
	}
	cout<<sum[c[n]]+mn[c[n]]<<'\n';
	return 0;
}
