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<algorithm>
#include<vector>
using namespace std;
const int N=1e5+5;
int n,m,id[N],fa[N],vis[N];
long long a[N],s[N],f[N];
vector<int>E[N];
bool cmp(int x,int y){
	return a[x]<a[y];
}
int find(int x){
	return x==fa[x]?x:fa[x]=find(fa[x]);
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i){
		scanf("%lld%lld",&a[i],&s[i]);
		f[i]=a[i]=max(a[i]-s[i],0ll);
		id[i]=fa[i]=i;
	}
	for(int i=1,x,y;i<=m;++i){
		scanf("%d%d",&x,&y);
		E[x].push_back(y);E[y].push_back(x);
	}
	sort(id+1,id+n+1,cmp);
	for(int i=1;i<=n;++i){
		int x=id[i];vis[x]=1;
		for(int y:E[x])
			if(vis[y]){
				y=find(y);
				if(y==x)continue;
				fa[y]=x;
				s[x]+=s[y];
				if(s[y]+f[y]>a[x])f[x]=min(f[x],f[y]);
				else f[x]=min(f[x],a[x]-s[y]);
			}
	}
	printf("%lld\n",s[find(1)]+f[find(1)]);
	return 0;
}
