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<queue>
#include<algorithm>
using namespace std;
#define N 100500
int vis[N],fa[N],s[N],v[N],n,m,a,b,head[N],cnt,is[N];
long long su[N],dp[N];
struct edge{int t,next;}ed[N*2];
void adde(int f,int t){ed[++cnt]=(edge){t,head[f]};head[f]=cnt;ed[++cnt]=(edge){f,head[t]};head[t]=cnt;}
int finds(int x){return fa[x]==x?x:fa[x]=finds(fa[x]);}
priority_queue<pair<long long,int> > tp;
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)scanf("%d%d",&s[i],&v[i]),tp.push(make_pair(-s[i]*2000000000ll+v[i],i));
	for(int i=1;i<=m;i++)scanf("%d%d",&a,&b),adde(a,b);
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<=n;i++)
	{
		int s1=tp.top().second;tp.pop();dp[s1]=1e17;is[s1]=1;
		for(int j=head[s1];j;j=ed[j].next)if(finds(ed[j].t)!=s1&&is[finds(ed[j].t)])vis[finds(ed[j].t)]=1,dp[s1]=min(dp[s1],max(1ll*s[s1],dp[finds(ed[j].t)]+v[s1])-su[finds(ed[j].t)]),su[s1]+=su[finds(ed[j].t)],fa[finds(ed[j].t)]=s1;
		dp[s1]+=su[s1];su[s1]+=v[s1];
		dp[s1]=min(dp[s1],max(s[s1],v[s1])+su[s1]-v[s1]);
	}
	printf("%lld\n",dp[finds(1)]);
}//