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<cstdio>
#include<vector>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long ll;
const int N=1e5+9;
int n,m,a[N],b[N],_[N],u,v,fa[N],t,d[N];
vector<int>e[N];
ll f[N],sz[N];
bool p[N];
ll min(ll x,ll y) {return x<y?x:y;}
ll max(ll x,ll y) {return x>y?x:y;}
bool cmp(int x,int y) {return a[x]<a[y];}
int gef(int x) {return fa[x]?fa[x]=gef(fa[x]):x;}
int main()
{
	scanf("%d%d",&n,&m);
	fo(i,1,n) scanf("%d%d",a+i,b+i), a[_[i]=i]=max(a[i]-b[i],0), sz[i]=b[i];
	fo(i,1,m) scanf("%d%d",&u,&v), e[u].push_back(v), e[v].push_back(u);
	sort(_+1,_+n+1,cmp);
	fo(i,1,n)
	{
		int x=_[i]; p[x]=1,t=0;
		for(int y:e[x]) if(p[y]&&gef(x)!=gef(y)) {fa[v=gef(y)]=x; sz[x]+=sz[d[++t]=v];}
		f[x]=sz[x]+a[x];
		for(int y;y=d[t];t--) f[x]=min(f[x],sz[x]-sz[y]+max(a[x],f[y]));
	}
	printf("%lld",f[gef(1)]);
}