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 N 100005
#define inf 1000000000000000000ll
using namespace std;
int i,j,k,l,n,m,x,y,last[N],to[2*N],Next[2*N],fa[N],id[N],V[N],v[N];
int tot,q[N];
long long A[N],s[N];
struct node {
	int x,y;
}a[N];
inline void add(int x,int y) {
	Next[++tot]=last[x]; last[x]=tot; to[tot]=y;
}
inline bool cmp(int x,int y) {
	return max(a[x].x-a[x].y,0)<max(a[y].x-a[y].y,0);
}
inline int get(int x) {
	if (fa[x]==x) return x;
	return fa[x]=get(fa[x]);
}
int main() {
	scanf("%d%d",&n,&m);
	for (i=1;i<=n;i++) scanf("%d%d",&a[i].x,&a[i].y);
	for (i=1;i<=m;i++) scanf("%d%d",&x,&y),add(x,y),add(y,x);
	for (i=1;i<=n;i++) fa[i]=id[i]=i,V[i]=0;
	sort(id+1,id+1+n,cmp);
	for (i=1;i<=n;i++) {
		int r=0;
		V[id[i]]=1;
		for (j=last[id[i]];j;j=Next[j]) {
			if (!V[to[j]]) continue;
			if (v[get(to[j])]!=id[i]) {
				q[++r]=get(to[j]),v[get(to[j])]=id[i];
			}
		}
		long long S=0,M=inf;
		if (!r) {
			A[id[i]]=max(a[id[i]].x,a[id[i]].y);
			s[id[i]]=a[id[i]].y;
			continue;
		}
		for (j=1;j<=r;j++) S+=s[q[j]];
		if (r==1) {
			s[id[i]]=S+a[id[i]].y;
			for (j=1;j<=r;j++) fa[q[j]]=id[i];
			int p=max(a[id[i]].x,a[id[i]].y);
			A[id[i]]=min(max(1ll*a[id[i]].x,1ll*a[id[i]].y+A[q[1]]),S+p);
			continue;
		}
		int p=max(a[id[i]].x,a[id[i]].y);
		for (j=1;j<=r;j++) M=min(M,S-s[q[j]]+max(1ll*a[id[i]].x,1ll*a[id[i]].y+A[q[j]]));
		M=min(M,S+p);
		for (j=1;j<=r;j++) fa[q[j]]=id[i];
		s[id[i]]=S+a[id[i]].y;
		A[id[i]]=M;
	}
	printf("%lld\n",A[id[n]]);
}