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;
typedef long long ll;
typedef pair<int,int> pii;
const int maxn=100010;
int a[maxn],b[maxn],c[maxn];
int par[maxn];
vector<int> init_g[maxn],g[maxn];
inline void init(int n){
	for(int i=1;i<=n;i++)
		par[i]=i;
}
int find(int x){
	if(par[x]==x) return x;
	return par[x]=find(par[x]);
}
ll sum[maxn],dp[maxn];
bool vis[maxn];
int ord[maxn];
inline bool cmp(const int& lhs,const int& rhs){
	return c[lhs]<c[rhs];
}
void dfs(int u){
	sum[u]=b[u];
	for(int i=0;i<(int)g[u].size();i++){
		int v=g[u][i];
		dfs(v);
		sum[u]+=sum[v];
	}
	dp[u]=sum[u]+c[u];
	for(int i=0;i<(int)g[u].size();i++){
		int v=g[u][i];
		dp[u]=min(dp[u],sum[u]-sum[v]+max((ll)c[u],dp[v]));
	}
}
int main(){
	int n,m,u,v;
	scanf("%d%d",&n,&m);
	init(n);
	for(int i=1;i<=n;i++){
		scanf("%d%d",a+i,b+i);
		c[i]=max(a[i]-b[i],0);
		ord[i]=i;
	}
	for(int i=0;i<m;i++){
		scanf("%d%d",&u,&v);
		init_g[u].push_back(v);
		init_g[v].push_back(u);
	}
	sort(ord+1,ord+1+n,cmp);
	for(int i=1;i<=n;i++){
		int u=ord[i];
		vis[u]=true;
		for(int j=0;j<(int)init_g[u].size();j++){
			int v=init_g[u][j];
			if(vis[v]){
				v=find(v);
				if(u!=v){
					g[u].push_back(v);
					par[v]=u;
				}
			}
		}
	}
	dfs(ord[n]);
	printf("%lld\n",dp[ord[n]]);
	return 0;
}