#include<bits/stdc++.h>
typedef long long LL;
using namespace std;
const int N=1e5+5;
int n,m,fa[N],p[N];
LL sum[N],f[N],a[N],b[N];
int head[N],ce;
bool vis[N];
struct edge{int to,nxt;} e[N<<1];
bool cmp(int x,int y) {return a[x]<a[y];}
void adde(int x,int y) {e[++ce].to=y,e[ce].nxt=head[x],head[x]=ce;}
int find(int x)
{
	if(x==fa[x]) return x;
	return fa[x]=find(fa[x]);
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)
	{
		scanf("%lld%lld",&a[i],&b[i]);
		a[i]=max(a[i]-b[i],0ll);
		p[i]=fa[i]=i,f[i]=a[i],sum[i]=b[i];
	}
	for(int i=1;i<=m;++i)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		adde(x,y); adde(y,x);
	}
	sort(p+1,p+n+1,cmp);
	for(int i=1;i<=n;++i)
	{
		int u=p[i]; vis[u]=1;
		for(int j=head[u];j;j=e[j].nxt)
		{
			int v=e[j].to;
			if(!vis[v]) continue;
			int x=find(u),y=find(v);
			if(x!=y)
			{
				fa[y]=x,sum[x]+=sum[y];
				f[x]=min(f[x],max(f[y],a[x]-sum[y]));
			}
		}
	}
	printf("%lld",f[find(1)]+sum[find(1)]);
	return 0;
}
### Prompt

Your challenge is to write a cpp solution to the following problem:

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

### Response

```cpp
#include<bits/stdc++.h>
typedef long long LL;
using namespace std;
const int N=1e5+5;
int n,m,fa[N],p[N];
LL sum[N],f[N],a[N],b[N];
int head[N],ce;
bool vis[N];
struct edge{int to,nxt;} e[N<<1];
bool cmp(int x,int y) {return a[x]<a[y];}
void adde(int x,int y) {e[++ce].to=y,e[ce].nxt=head[x],head[x]=ce;}
int find(int x)
{
	if(x==fa[x]) return x;
	return fa[x]=find(fa[x]);
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;++i)
	{
		scanf("%lld%lld",&a[i],&b[i]);
		a[i]=max(a[i]-b[i],0ll);
		p[i]=fa[i]=i,f[i]=a[i],sum[i]=b[i];
	}
	for(int i=1;i<=m;++i)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		adde(x,y); adde(y,x);
	}
	sort(p+1,p+n+1,cmp);
	for(int i=1;i<=n;++i)
	{
		int u=p[i]; vis[u]=1;
		for(int j=head[u];j;j=e[j].nxt)
		{
			int v=e[j].to;
			if(!vis[v]) continue;
			int x=find(u),y=find(v);
			if(x!=y)
			{
				fa[y]=x,sum[x]+=sum[y];
				f[x]=min(f[x],max(f[y],a[x]-sum[y]));
			}
		}
	}
	printf("%lld",f[find(1)]+sum[find(1)]);
	return 0;
}
```