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<cstring>
#include<algorithm>
#include<vector>
using namespace std;
struct bian
{
	int nxt,to;
}bi[200020];
int n,m,head[100010],num,a[100010],b[100010],c[100010],f[100010],p[100010];
long long sum[100010],dp[100010];
vector<int>E[100010];
inline void add(int from,int to){bi[++num]=bian{head[from],to};head[from]=num;}
bool cmp(int x,int y){return c[x]<c[y];}
int find(int x){if(x!=f[x])return f[x]=find(f[x]);return x;}
void dfs(int v)
{
	sum[v]=b[v];
	if(!E[v].size())
	{
		dp[v]=max(a[v],b[v]);
		return ;
	}
	for(auto u:E[v])
	{
		dfs(u);
		sum[v]+=sum[u];
	}
	dp[v]=0x3f3f3f3f3f3f3f3f;
	for(auto u:E[v])dp[v]=min(dp[v],sum[v]-sum[u]+max(1ll*c[v],dp[u]));
}
int main()
{
	scanf("%d %d",&n,&m);
	for(int i=1;i<=n;++i)scanf("%d %d",&a[i],&b[i]),c[i]=max(a[i]-b[i],0);
	int x,y;
	for(int i=1;i<=m;++i)
	{
		scanf("%d %d",&x,&y);
		add(x,y),add(y,x);
	}
	for(int i=1;i<=n;++i)p[i]=i;
	sort(p+1,p+1+n,cmp);
	for(int i=1;i<=n;++i)
	{
		x=p[i];
		f[x]=x;
		for(int j=head[x],u;j;j=bi[j].nxt)
		{
			u=find(bi[j].to);
			if(u&&u!=x)
			{
				f[u]=x;
				E[x].push_back(u);
			}
		}
	}
	dfs(p[n]);
	printf("%lld",dp[p[n]]);
	return 0;
}
