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;
#define ll long long
#define REP(i,a,b) for(int i=(a),_end_=(b);i<=_end_;i++)
#define DREP(i,a,b) for(int i=(a),_end_=(b);i>=_end_;i--)
#define EREP(i,u) for(int i=start[u];i;i=e[i].next)
#define fi first
#define se second
#define mkr(a,b) make_pair(a,b)
#define SZ(A) ((int)A.size())
template<class T>inline void chkmin(T &a,T b){ if(a>b)a=b;}
template<class T>inline void chkmax(T &a,T b){ if(a<b)a=b;}
inline void myassert(bool a,string s){ if(!a)cerr<<s<<endl,exit(0);}
inline int read()
{
	int s=0,f=1;char ch=getchar();
	while(!isdigit(ch) && ch!='-')ch=getchar();
	if(ch=='-')ch=getchar(),f=-1;
	while(isdigit(ch))s=s*10+ch-'0',ch=getchar();
	return ~f?s:-s;
}

const int maxn=1e5+20;

int n,m;
vector<int>ed[maxn];
int a[maxn],b[maxn];
int fa[maxn],vis[maxn],id[maxn];
ll sum[maxn],mn[maxn];

inline void init()
{
	n=read();m=read();
	REP(i,1,n)
	{
		a[i]=read();b[i]=read();
		a[i]=max(a[i]-b[i],0);
		fa[i]=id[i]=i;
		sum[i]=b[i]; mn[i]=a[i];
	}
	REP(i,1,m)
	{
		int x=read(),y=read();
		ed[x].push_back(y);
		ed[y].push_back(x);
	}
	sort(id+1,id+n+1,[](int x,int y){ return a[x]<a[y];});
}

int fin(int x){ return x==fa[x]?x:fa[x]=fin(fa[x]);}

inline void doing()
{
	REP(i,1,n)
	{
		int x=id[i]; vis[x]=1;
		for(int y:ed[x])
			if(vis[y])
			{
				int u=fin(x),v=fin(y);
				if(u==v)continue;
				fa[v]=u;
				mn[u]=min(max(mn[v]-sum[u],mn[u]),max(a[u]-sum[v],mn[v]));
				sum[u]+=sum[v];
			}
	}
	printf("%lld\n",sum[id[n]]+mn[id[n]]);
}

int main()
{
	init();
	doing();
	return 0;
}
