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 y1 dmytxdy
#define fi first
#define se second
#define mp make_pair

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef long double ld;

template <typename T> bool chkmin(T &x,T y){return x>y?x=y,1:0;}
template <typename T> bool chkmax(T &x,T y){return x<y?x=y,1:0;}

int readint(){
	int x=0,f=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

int n,m;
ll a[100005],b[100005],siz[100005],ans[100005],f[100005];
pll p[100005];
vector<int> adj[100005];

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

int main(){
	n=readint(); m=readint();
	for(int i=1;i<=n;i++) a[i]=readint(),b[i]=readint(),chkmax(a[i],b[i]);
	int x,y;
	for(int i=1;i<=m;i++){
		x=readint(); y=readint();
		if(x>y) swap(x,y);
		if(a[x]-b[x]>a[y]-b[y]) adj[x].push_back(y);
		else adj[y].push_back(x);
	}
	for(int i=1;i<=n;i++) p[i]=mp(a[i]-b[i],i);
	for(int i=1;i<=n;i++) f[i]=i,siz[i]=b[i],ans[i]=a[i]-b[i];
	sort(p+1,p+n+1);
	for(int i=1;i<=n;i++){
		int u=p[i].se;
		for(auto x:adj[u]){
			int fx=getf(x);
			if(fx==u) continue;
			chkmin(ans[u],max(a[u]-b[u]-siz[fx],ans[fx]));
			siz[u]+=siz[fx];
			f[fx]=u;
		}
	}
	printf("%lld\n",ans[getf(1)]+siz[getf(1)]);
	return 0;
}