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<algorithm>
#include<cstdio>
#define N 100005
using namespace std;
int a[N],b[N],c[N],d[N],e[N],g[N],f[N],m,n;long long h[N],x;
inline bool cmp(int u,int v){return g[a[u]]==g[a[v]]?a[u]<a[v]:g[a[u]]<g[a[v]];}
int find(int u)
{
	if(!f[u])return u;
	int v=find(f[u]);
	return e[u]=max(e[u],e[f[u]]),f[u]=v;
}
int main()
{
	scanf("%d%d",&n,&m),x=0x7fffffffffffffffll;
	for(int i=1;i<=n;i++)scanf("%d%lld",g+i,h+i),g[i]=g[i]<h[i]?0:g[i]-h[i];
	for(int i=1;i<=m;i++)scanf("%d%d",a+i,b+i),c[i]=i,g[a[i]]<g[b[i]]||g[a[i]]==g[b[i]]&&a[i]<b[i]?a[i]^=b[i]^=a[i]^=b[i]:0;
	sort(c+1,c+m+1,cmp);
	for(int i=1,u;i<=m;i++)if(a[c[i]]!=(u=find(b[c[i]])))e[u]=max((long long)e[u],g[a[c[i]]]-h[u]),f[u]=a[c[i]],h[a[c[i]]]+=h[u];
	for(int i=1,u;i<=n;i++)u=find(i),x=min(x,max(g[i],e[i])+h[u]);
	return 0*printf("%lld\n",x);
}