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
using namespace std;
const int N=3e5+99;
struct node{int id;ll a,b;}p[N];
int n,m,fa[N],vis[N],cur[N];
ll a[N],b[N],sum[N],ans[N];
vector<int>g[N];
bool cmp(node a,node b){return a.a<b.a;}
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		scanf("%d%d",&a[i],&b[i]);
		a[i]=max(a[i],b[i]);
		p[i]=(node){i,a[i],b[i]};
		sum[i]=b[i];
	}
	for(int i=1,x,y;i<=m;i++)scanf("%d%d",&x,&y),g[x].push_back(y),g[y].push_back(x);
	sort(p+1,p+n+1,cmp);
	for(int i=1;i<=n;i++)fa[i]=i,ans[i]=1e18;
	for(int i=1,j;i<=n;i=j+1)
	{
		j=i;while(p[j+1].a==p[i].a)j++;
		for(int k=i;k<=j;k++)vis[p[k].id]=1;
		for(int k=i;k<=j;k++)
		{
			int x=p[k].id;
			for(int t=0;t<g[x].size();t++)
			{
				int to=find(g[x][t]);
				if(!vis[to])continue;
				cur[g[x][t]]=to;
			}
		}
		for(int k=i;k<=j;k++)
		{
			int x=p[k].id;
			for(int t=0;t<g[x].size();t++)
			{
				int to=find(g[x][t]);
				if(!vis[to]||find(x)==to)continue;
				int y=find(x),z=find(to);
				fa[z]=y;
				sum[y]+=sum[to];
			}
		}
		for(int k=i;k<=j;k++)
		{
			int x=p[k].id;
			for(int t=0;t<g[x].size();t++)
			{
				int to=cur[g[x][t]];
				if(!vis[to]||i==vis[to])continue;
				ll now=find(x),cst=max(a[x],ans[to]+b[x])+sum[now]-sum[to]-b[x];
				ans[now]=min(ans[now],cst);
			}
			int y=find(x);
			ans[y]=min(ans[y],a[y]+sum[y]-b[x]);
		}
	}
	cout<<ans[find(1)]<<endl;
}