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=1e5+5;
int n,m,a[N],b[N],u,v,x,i,g[N],id[N];
vector<int>e[N];
long long f[N],su[N];
int gfa(int x){return g[x]==x?x:g[x]=gfa(g[x]);}
inline bool cmp(const int&x,const int&y){return a[x]-b[x]<a[y]-b[y];}
int main(){
	ios::sync_with_stdio(0);cin.tie(0);
	cin>>n>>m;for(i=1;i<=n;++i)cin>>a[i]>>b[i],g[i]=id[i]=i;
	for(i=1;i<=m;++i)cin>>u>>v,e[u].push_back(v),e[v].push_back(u);
	sort(id+1,id+n+1,cmp);
	for(i=1;i<=n;++i){
		u=id[i];su[u]=b[u];f[u]=1ll<<60;
		for(int v:e[u])if(su[v]){
			x=gfa(v);if(x==u)continue;
			f[u]=min(f[u],max(f[x],a[u]-su[x]-b[u]));su[u]+=su[x];g[x]=u;
		}
		if(su[u]==b[u])f[u]=max(a[u]-b[u],0);
	}
	cout<<f[u]+su[u]<<endl;
	return 0;
}