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;
const int N=1e5+10;
typedef vector<int> vi;
#define pb push_back
typedef long long ll;

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,m,a[N],b[N],fa[N],id[N];
ll s[N],f[N];
vi E[N];
bool vis[N];

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

bool cmp(int x,int y) {
	return a[x]<a[y];
}

int main() {
	cin>>n>>m;
	for(int i=1;i<=n;i++) {
		a[i]=gi(),b[i]=gi();
		a[i]=max(a[i]-b[i],0);
		id[i]=fa[i]=i;s[i]=b[i];f[i]=a[i]+b[i];
	}
	for(int i=1,u,v;i<=m;i++) u=gi(),v=gi(),E[u].pb(v),E[v].pb(u);
	sort(id+1,id+n+1,cmp);
	for(int i=1;i<=n;i++) {
		int u=id[i];vis[u]=1;
		for(auto v:E[u])
			if(vis[v]) {
				int x=u,y=getf(v);
				if(x!=y) {
					fa[y]=x;f[x]=min(s[x]+max((ll)a[x],f[y]),s[y]+f[x]);s[x]+=s[y];
				}
			}
	}
	cout<<f[getf(1)];
	return 0;
}
