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;
#define MN 100000
struct edge{int x,y,w;}e[MN+5];
bool cmp(const edge&a,const edge&b){return a.w<b.w;}
int a[MN+5],b[MN+5],f[MN+5];long long s[MN+5],ss[MN+5];
int gf(int k){return f[k]?f[k]=gf(f[k]):k;}
int main()
{
	int n,m,i;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;++i)scanf("%d%d",&a[i],&b[i]),s[i]=max(a[i],b[i]),ss[i]=b[i];
	for(i=1;i<=m;++i)scanf("%d%d",&e[i].x,&e[i].y),e[i].w=max(a[e[i].x]-b[e[i].x],a[e[i].y]-b[e[i].y]);
	sort(e+1,e+m+1,cmp);
	for(i=1;i<=m;++i)if((e[i].x=gf(e[i].x))!=(e[i].y=gf(e[i].y)))
	{
		s[e[i].x]=min(max(s[e[i].x],1LL*e[i].w)+ss[e[i].y],max(s[e[i].y],1LL*e[i].w)+ss[e[i].x]);
		ss[e[i].x]+=ss[e[i].y];
		f[e[i].y]=e[i].x;
	}
	printf("%lld",s[gf(1)]);
}