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>
typedef long long ll;
ll gi(){
	ll x=0,f=1;
	char ch=getchar();
	while(!isdigit(ch))f^=ch=='-',ch=getchar();
	while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
	return f?x:-x;
}
int fa[100010],A[100010],B[100010],s[100010];
ll dp[100010],sb[100010];
std::vector<int>G[100010];
int hd(int x){return fa[x]==x?x:fa[x]=hd(fa[x]);}
bool yes[100010];
int main(){
#ifdef XZZSB
	freopen("in.in","r",stdin);
	freopen("out.out","w",stdout);
#endif
	int n=gi(),m=gi(),a,b;
	for(int i=1;i<=n;++i)A[i]=gi(),B[i]=sb[i]=gi(),fa[i]=i,dp[i]=std::max(0,A[i]-B[i]),s[i]=i;
	std::sort(s+1,s+n+1,[&](int x,int y){return dp[x]<dp[y];});
	while(m--)a=gi(),b=gi(),G[a].push_back(b),G[b].push_back(a);
	for(int _=1,x;_<=n;++_){
		x=s[_];yes[x]=1;
		for(int i:G[x])
			if(hd(i)!=hd(x)&&yes[i]){
				int s=hd(i);
				fa[s]=x;
				sb[x]+=sb[s];
				dp[x]=std::min(dp[x],dp[s]+std::max(0ll,A[x]-B[x]-sb[s]-dp[s]));
			}
	}
	printf("%lld\n",dp[hd(1)]+sb[hd(1)]);
	return 0;
}
