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
/****************************************************************
*	Author: huhao
*	Email: 826538400@qq.com
*	Create time: 2020-02-09 19:16:44
****************************************************************/
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define fr(i,a,b) for(int i=(a),end_##i=(b);i<=end_##i;i++)
#define fd(i,a,b) for(int i=(a),end_##i=(b);i>=end_##i;i--)
int read()
{
	int r=0,t=1,c=getchar();
	while(c<'0'||c>'9')
	{
		t=c=='-'?-1:1;
		c=getchar();
	}
	while(c>='0'&&c<='9')
	{
		r=(r<<3)+(r<<1)+(c^48);
		c=getchar();
	}
	return r*t;
}
#define i64 long long
#include<vector>
const int N=100010;
i64 n,m,a[N],b[N],f[N],F[N],rk[N],s[N];
int getf(int u){ return u==F[u]?u:F[u]=getf(F[u]); }
std::vector<int> E[N],e[N];
int cmp(int x,int y){ return a[x]-b[x]>a[y]-b[y]; }
void dfs(int u)
{
	s[u]=b[u]; f[u]=a[u]-b[u];
	fr(i,0,e[u].size()-1)
	{
		dfs(e[u][i]);
		s[u]+=s[e[u][i]]; f[u]=std::min(f[u],std::max(a[u]-b[u]-s[e[u][i]],f[e[u][i]]));
	}
}
int main()
{
	n=read(); m=read();
	fr(i,1,n){ a[i]=read(); b[i]=read(); a[i]=std::max(a[i],b[i]); }
	fr(i,1,m)
	{
		int u=read(),v=read();
		E[u].push_back(v); E[v].push_back(u);
	}
	fr(i,1,n) rk[i]=i;
	std::sort(rk+1,rk+n+1,cmp);
	fd(I,n,1)
	{
		int i=rk[I]; F[i]=i;
		fr(j,0,E[i].size()-1) if(F[E[i][j]])
		{
			int p=getf(E[i][j]);
			if(p!=i){ F[p]=i; e[i].push_back(p); }
		}
	}
	dfs(rk[1]);
	printf("%lld\n",f[rk[1]]+s[rk[1]]);
	return 0;
}