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<cctype>
#include<vector>
#define pb push_back
#define forall(x,i) for (int i=0;i<x.size();i++)
using namespace std;
#define G getchar()
int read()
{
	int x=0; char ch=G;
	for (;!isdigit(ch);ch=G);
	for (;isdigit(ch);ch=G) x=(x<<3)+(x<<1)+(ch^48);
	return x;
}
#undef G
typedef long long ll;

int n,m;
int A[100010],B[100010],C[100010];
int id[100010];
bool cmp(int x,int y){return C[x]<C[y];}
struct Edge{
	int to,nxt;
}edge[200010];
int cnt=1,last[100010];
inline void addedge(int x,int y){
	edge[++cnt]=(Edge){y,last[x]},last[x]=cnt;
	edge[++cnt]=(Edge){x,last[y]},last[y]=cnt;
}
ll S[100010],f[100010];
bool vis[100010];
int fat[100010];
int find(int x){return fat[x]?fat[x]=find(fat[x]):x;}
int main()
{
	n=read(),m=read();
	for (int i=1;i<=n;i++){
		A[i]=read(),B[i]=read();
		A[i]=max(A[i],B[i]);
		C[i]=A[i]-B[i];
		id[i]=i;
	}
	for (int i=1;i<=m;i++) addedge(read(),read());
	sort(id+1,id+n+1,cmp);
	for (int i=1;i<=n;i++){
		static int e[100010],top; top=0;
		int x=id[i]; vis[x]=1;
		S[x]=B[x];
		for (int j=last[x],y;j;j=edge[j].nxt){
			y=edge[j].to;
			if (!vis[y]||x==find(y)) continue;
			int k=find(y); S[x]+=S[k]; fat[k]=x;
			e[++top]=k;
		}
		f[x]=S[x]+C[x];
		for (int j=1;j<=top;j++){
			int k=e[j];
			f[x]=min(f[x],S[x]-S[k]+max(0LL+C[x],f[k]));
		}
	}
	printf("%lld\n",f[id[n]]);
	return 0;
}