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
int n,m,x,y,L,R,M,T,fl;
int f[100005],vis[100005],a[100005],b[100005],rt[100005];
vector <int> e[100005];
int g[500005],lc[500005],rc[500005];
LL sum,w[100005];
int find(int x){
	return f[x]==x?x:f[x]=find(f[x]);
}
int he(int u,int v){
	if (!u||!v) return u+v;
	if (a[g[u]]-b[g[u]]>a[g[v]]-b[g[v]]) swap(u,v);
	lc[u]=he(lc[u],v);
	swap(lc[u],rc[u]);
	return u;
}
int main(){
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;++i) scanf("%d%d",&a[i],&b[i]),sum+=b[i];
	for (int i=1;i<=m;++i){
		scanf("%d%d",&x,&y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	L=0; R=1000000000;
	while (L<R){
		M=(L+R)/2; T=0; fl=0;
		for (int i=1;i<=n;++i)
			{f[i]=i; w[i]=M; vis[i]=0;}
		for (int i=1;i<=n;++i)
		if (a[i]<=b[i]+M&&!vis[i]){
			g[++T]=i; lc[T]=rc[T]=0; rt[i]=T;
			while (rt[i]){
				x=g[rt[i]];
				if (a[x]-b[x]>w[i]) break;
				rt[i]=he(lc[rt[i]],rc[rt[i]]);
				if (vis[x]) continue;
				w[i]+=b[x]; f[x]=i; vis[x]=1;
				for (unsigned j=0;j<e[x].size();++j){
					y=find(e[x][j]);
					if (y==i) continue;
					if (!vis[y]){
						g[++T]=y; lc[T]=rc[T]=0;
						rt[i]=he(rt[i],T);
					}else{
						rt[i]=he(rt[i],rt[y]);
						f[y]=i; w[i]+=w[y]-M;
					}
				}
				if (w[i]==sum+M) {fl=1; break;}
			}
			if (fl) break;
		}
		if (fl) R=M; else L=M+1;
	}
	printf("%lld\n",sum+L);
	return 0;
}
