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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define maxn 100005
#define ll long long 
using namespace std;

int n,m,i,j,k,a[maxn],b[maxn],c[maxn],id[maxn];
int em,e[maxn*2],nx[maxn*2],ls[maxn];
int cmp(int i,int j){return c[i]<c[j];}

void insert(int x,int y){
	em++; e[em]=y; nx[em]=ls[x]; ls[x]=em;
	em++; e[em]=x; nx[em]=ls[y]; ls[y]=em;
}

int fa[maxn]; ll g[maxn],f[maxn];
int father(int x){return (fa[x]==x)?x:fa[x]=father(fa[x]);}

int main(){
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++) scanf("%d%d",&a[i],&b[i]),c[i]=a[i]-b[i];
	for(i=1;i<=m;i++) scanf("%d%d",&j,&k),insert(j,k);
	for(i=1;i<=n;i++) id[i]=i;sort(id+1,id+1+n,cmp);
	for(int now=1;now<=n;now++){int x=id[now];
		fa[x]=x,f[x]=max(0,c[x]),g[x]=b[x];
		for(i=ls[x];i;i=nx[i]) if (fa[e[i]]){int y=e[i];
			if (father(x)!=father(y)){
				y=fa[y];
				f[x]=min(f[x],max(f[y],c[x]-g[y]));
				g[x]+=g[y],fa[y]=x;
			}
		}
	}
	printf("%lld\n",f[father(1)]+g[father(1)]);
}