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>
#define rep(i,x,y) for (int i=(x);i<=(y);i++)
#define ll long long
#define inf 1000000001
using namespace std;
ll read(){
	char ch=getchar();ll x=0;int op=1;
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') op=-1;
	for (;isdigit(ch);ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
	return x*op;
}
#define N 100005
int n,m,cnt,head[N],a[N],b[N],fa[N],id[N],vis[N];ll s1[N],s2[N];
struct edge{int to,nxt;}e[N<<1];
void adde(int x,int y){e[++cnt].to=y;e[cnt].nxt=head[x];head[x]=cnt;}
void ins(int x,int y){adde(x,y);adde(y,x);}
const bool cmp(const int x,const int y){return a[x]<a[y];}
int getfa(int x){return x==fa[x]?x:fa[x]=getfa(fa[x]);}
int main(){
	n=read(),m=read();
	rep (i,1,n){a[i]=read(),b[i]=read();a[i]=max(a[i]-b[i],0);}
	rep (i,1,m){int x=read(),y=read();ins(x,y);}
	rep (i,1,n) fa[i]=id[i]=i,s1[i]=a[i],s2[i]=b[i];
	sort(&id[1],&id[n+1],cmp);
	rep (i,1,n){
		int x=id[i];vis[x]=1;
		for (int j=head[x],y;j;j=e[j].nxt){
			y=e[j].to;if (!vis[y]) continue;
			int fx=getfa(x),fy=getfa(y);
			if (fx!=fy){
				fa[fy]=fx;s2[fx]+=s2[fy];
				s1[fx]=min(s1[fx],s1[fy]+max(0ll,a[x]-s1[fy]-s2[fy]));
			}
		}
	}
	int x=getfa(1);
	cout<<s1[x]+s2[x]<<'\n';
	return 0;
}