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 ll long long

vector<int> v[100010],g[100010];

ll sum[100010],f[100010];
int A[100010],B[100010],h[100010],from[100010],id[100010],n,m;

inline int rd()
{
	int x=0;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar());
	for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

inline bool cmp(const int &x,const int &y) { return A[x]<A[y]; }

inline int find(int x) { return (x==from[x])?x:from[x]=find(from[x]); }

int main()
{
	n=rd();m=rd();
	for (int i=1;i<=n;i++) A[i]=rd(),B[i]=rd(),A[i]=max(0,A[i]-B[i]);
	for (int i=1;i<=m;i++)
	{
		int x=rd(),y=rd();
		v[x].push_back(y);
		v[y].push_back(x);
	}
	for (int i=1;i<=n;i++) h[i]=i;
	sort(h+1,h+n+1,cmp);
	for (int i=1;i<=n;i++) from[i]=i,id[h[i]]=i;
	for (int i=1;i<=n;i++)
	{
		int x=h[i];
		for (int t:v[x]) if (id[t]<i&&find(t)!=x) g[x].push_back(find(t)),from[find(t)]=x;
	}
	for (int i=1;i<=n;i++)
	{
		int x=h[i];
		sum[x]=B[x];
		for (int t:g[x]) sum[x]+=sum[t];
		f[x]=sum[x]+A[x];
		for (int t:g[x]) f[x]=min(f[x],max(f[t],(ll)A[x])+sum[x]-sum[t]);
	}
	printf("%lld\n",f[h[n]]);
	return 0;
}