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> 
#define LL long long
#define clr(x,i) memset(x,i,sizeof(x))
using namespace std;
const int N=100007;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
vector<int> G[N];
int n,m,a[N],b[N],fa[N],rk[N],vis[N];
LL f[N],sum[N],ans;
bool cmp(int x,int y)
{
	return a[x]<a[y];
}
int getfa(int x)
{
	return fa[x]==x ? x : fa[x]=getfa(fa[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],0);
	}
	for(int i=1;i<=m;i++){
		int u=read(),v=read();
		G[u].push_back(v);G[v].push_back(u);
	}
	for(int i=1;i<=n;i++){
		rk[i]=fa[i]=i; f[i]=a[i]; sum[i]=b[i];
	}
	sort(rk+1,rk+n+1,cmp);
	for(int i=1;i<=n;i++)
	{
		int u=rk[i]; vis[u]=1;
		for(int j=0;j<G[u].size();j++)
		{
			int v=G[u][j];
			if(!vis[v]) continue;
			int x=getfa(u),y=getfa(v);
			if(x!=y){
				fa[y]=x; sum[x]+=sum[y];
//				f[x]=min(f[x],f[y]+max(f[y],a[x]-sum[y]));
				f[x]=min(f[x],max(f[y],a[x]-sum[y]));
			}
		}
	}
	int rt=getfa(1);
	printf("%lld\n",f[rt]+sum[rt]);
	return 0;
}
