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;
template <typename T> void read(T &t) {
	t=0; char ch=getchar(); int f=1;
	while (ch<'0'||ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	do { (t*=10)+=ch-'0'; ch=getchar(); } while ('0'<=ch&&ch<='9'); t*=f;
}
typedef long long ll;
const int maxn=(1e5)+10;
int n,m,fa[maxn],d[maxn];
int a[maxn],b[maxn];
ll f[maxn],g[maxn];
bool vis[maxn];
int tot,nxt[maxn*2],head[maxn],to[maxn*2];
bool cmp(int x,int y) { return a[x]<a[y]; }
void add(int x,int y) {
	tot++; nxt[tot]=head[x];
	head[x]=tot; to[tot]=y;
}
int find(int x) {
	if (fa[x]==x) return x;
	return fa[x]=find(fa[x]);
}
void merge(int x,int y) {
	y=find(y);
	if (x==y) return;
	fa[y]=x;
	g[x]+=g[y];
	f[x]=min(f[x],max(f[y],a[x]-g[y]));
}
int main() {
//	freopen("1.txt","r",stdin);
	read(n); read(m);
	for (int i=1;i<=n;i++) {
		read(a[i]); read(b[i]);
		a[i]=max(0,a[i]-b[i]);
		fa[i]=i; d[i]=i;
		f[i]=a[i]; g[i]=b[i];
	}
	sort(d+1,d+n+1,cmp);
	int x,y;
	for (int i=1;i<=m;i++) {
		read(x); read(y);
		add(x,y); add(y,x);
	}
	for (int i=1;i<=n;i++) {
		x=d[i];
		vis[x]=1;
		for (int j=head[x];j;j=nxt[j]) {
			y=to[j];
			if (vis[y]) merge(x,y);
		}
	}
	printf("%lld\n",f[find(1)]+g[find(1)]);
	return 0;
}
/*
  0. Enough array size? Enough array size? Enough array size? Interger overflow?
  
  1. Think TWICE, Code ONCE!
  Are there any counterexamples to your algo?
    
  2. Be careful about the BOUNDARIES!
  N=1? P=1? Something about 0?
    
  3. Do not make STUPID MISTAKES!
  Time complexity? Memory usage? Precision error?
*/