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<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
using namespace std;
typedef long long ll;
#define N 500002
struct edge{int to,nxxt;}e[N<<1];
vector<int>v[N];
int head[N],cnt=1,m,n,a[N],b[N],w[N],id[N],f[N];
bool vis[N];
ll sum[N],dp[N];
bool cmp(int x,int y){return w[x]<w[y];}
inline void ins(int u,int v)
{e[cnt].to=v;e[cnt].nxxt=head[u];head[u]=cnt++;}
int getroot(int t){return f[t]==0?t:f[t]=getroot(f[t]);}
void dfs(int te)
{sum[te]=b[te];
	for(int i=0;i<v[te].size();i++)
	{
		int j=v[te][i];dfs(j);sum[te]+=sum[j];
	}dp[te]=b[te]+w[te];if(!v[te].empty())dp[te]=1e18;
	for(int i=0;i<v[te].size();i++)
	{int j=v[te][i];
		dp[te]=min(dp[te],sum[te]-sum[j]+max(1ll*w[te],dp[j]));
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)scanf("%d%d",&a[i],&b[i]),w[i]=max(0,a[i]-b[i]);
	for(int i=1;i<=m;i++)
	{
		int x,y;scanf("%d%d",&x,&y);
		ins(x,y),ins(y,x);
	}
	for(int i=1;i<=n;i++)id[i]=i;
	sort(id+1,id+1+n,cmp);
	for(int i=1;i<=n;i++)
	{
		int x=id[i];
		for(int j=head[x];j;j=e[j].nxxt)
		{
			int y=e[j].to;
			int yy=getroot(y);
			if(x!=yy&&vis[yy])
			{
				f[yy]=x;
				v[x].push_back(yy);
			}
		}vis[x]=1;
	}dfs(id[n]);
	printf("%lld\n",dp[id[n]]);
}
