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 maxn=2e5;
int n,m,u,v,fa[maxn];
long long a[maxn],b[maxn],c[maxn],f[maxn],sum[maxn],id[maxn];
bool vis[maxn];
vector <int> to[maxn];
bool cmp(int x,int y){return c[x]<c[y];}
int get_fa(int now){return fa[now]=((fa[now]==now)?now:get_fa(fa[now]));}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		scanf("%lld%lld",&a[i],&b[i]);
		c[i]=max(a[i]-b[i],0ll);
		id[i]=i;fa[i]=i;sum[i]=b[i];f[i]=c[i];
	}
	for(int i=1;i<=m;i++){
		scanf("%d%d",&u,&v);
		to[u].push_back(v);
		to[v].push_back(u);
	}
	sort(id+1,id+n+1,cmp);
	for(int i=1;i<=n;i++){
		int now=id[i],siz;
		vis[now]=true;siz=to[now].size();
		for(int j=0;j<siz;j++){
			int t=to[now][j];
			if(!vis[t]) continue;
			t=get_fa(t);
			if(t==now) continue;
			fa[t]=now;
			sum[now]+=sum[t];
			f[now]=min(f[now],max(f[t],c[now]-sum[t]));
		}
	}
	printf("%lld\n",f[id[n]]+sum[id[n]]);
	return 0;
}