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 
#define N 300000
#define INF 0x3f3f3f3f3f3f3f3fLL

LL n,m,x,y,a[N],b[N],c[N],d[N],f[N],fa[N],sum[N],tot,head[N];
bool vis[N];
struct edge{LL v,nxt;}e[N];
vector<LL> vec[N];

void add(LL x,LL y){e[++tot].v=y; e[tot].nxt=head[x]; head[x]=tot;}

bool cmp(LL x,LL y){return c[x]<c[y];}

LL gfa(LL x){return !fa[x]?x:fa[x]=gfa(fa[x]);}

void dp(LL u){
	sum[u]=b[u]; f[u]=INF;
	for (LL i=0,v;i<(LL)vec[u].size();++i){dp(v=vec[u][i]); sum[u]+=sum[v];}
	if (!vec[u].size()){f[u]=b[u]+c[u]; return;}
	for (LL i=0,v;i<(LL)vec[u].size();++i){
		v=vec[u][i];
		f[u]=min(f[u],sum[u]-sum[v]+max(c[u],f[v]));
	}
}

int main(){
	scanf("%lld%lld",&n,&m);
	for (LL i=1;i<=n;++i){
		scanf("%lld%lld",&a[i],&b[i]);
		a[i]=max(a[i],b[i]);
		c[i]=a[i]-b[i];
		d[i]=i;
	}
	for (LL i=1;i<=m;++i){
		scanf("%lld%lld",&x,&y);
		add(x,y); add(y,x);
	}
	sort(d+1,d+n+1,cmp);
	for (LL i=1;i<=n;++i){
		LL x=d[i];
		for (LL j=head[x];j;j=e[j].nxt)
			if (vis[e[j].v]){
				LL y=gfa(e[j].v);
				if (x!=y){fa[y]=x; vec[x].push_back(y);}
			}
		vis[x]=1;
	}
	dp(d[n]);
	printf("%lld\n",f[d[n]]);
	
	return 0;
}
			