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>
using namespace std;
typedef long long ll;
int n,m,fa[100010],a[100010],b[100010];
vector<int> v[100010];
ll s[100010];
inline int getf(int x)
{
	return x==fa[x]?x:fa[x]=getf(fa[x]);
}
inline void merge(int x,int y)
{
	x=getf(x);y=getf(y);
	fa[x]=y;
	if(x!=y)s[y]+=s[x];
}
pair<int,int> p[100010];
bool flag[100010];
ll f[100010];
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&a[i],&b[i]);
		p[i]=make_pair(a[i]-b[i],i);
	}
	sort(p+1,p+n+1);
	for(int i=1,a,b;i<=m;i++)
	{
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
	}
	for(int i=1;i<=n;i++)fa[i]=i;
	for(int i=1;i<=n;i++)
	{
		int pos=p[i].second;
		f[pos]=max(a[pos]-b[pos],0);
		s[pos]=b[pos];
		for(int &x:v[pos])
		{
			if(flag[x])
			{
				int t=getf(x);
				f[pos]=min(f[pos],max(f[t],a[pos]-b[pos]-s[t]));
			}
		}
		for(int &x:v[pos])
			if(flag[x])merge(x,pos);
		flag[pos]=1;
	}
	printf("%lld\n",f[p[n].second]+s[p[n].second]);
	return 0;
}