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 

const int maxn=500010, inf=1e9;
struct edge{int too, pre;}e[maxn];
int n, m, x, y, tot;
int fa[maxn], a[maxn], b[maxn], last[maxn], pos[maxn];
ll ad[maxn], sum[maxn];
bool v[maxn];
template<typename T>
inline void read(T &k)
{
	int f=1; k=0; char c=getchar();
	while(c<'0' || c>'9') c=='-'&&(f=-1), c=getchar();
	while(c<='9' && c>='0') k=k*10+c-'0', c=getchar();
	k*=f;
}
inline void add(int x, int y){e[++tot]=(edge){y, last[x]}; last[x]=tot;}
inline bool cmp(int x, int y){return a[x]<a[y];}
int gf(int x){return fa[x]==x?x:fa[x]=gf(fa[x]);}
int main()
{
	read(n); read(m);
	for(int i=1;i<=n;i++) read(a[i]), read(b[i]), a[i]=max(a[i]-b[i], 0);
	for(int i=1;i<=m;i++) read(x), read(y), add(x, y), add(y, x);
	for(int i=1;i<=n;i++) pos[i]=i, fa[i]=i, ad[i]=a[i], sum[i]=b[i];
	sort(pos+1, pos+1+n, cmp);
	for(int i=1;i<=n;i++)
	{
		int x=pos[i]; v[x]=1;
		for(int j=last[x], too;j;j=e[j].pre)
		if(v[too=e[j].too])
		{
			int fx=gf(x), fy=gf(too);
			if(fx==fy) continue;
			sum[fx]+=sum[fy]; fa[fy]=fx;
			ad[fx]=min(ad[fx], ad[fy]+max(0ll, a[x]-ad[fy]-sum[fy]));
		}
	}
	printf("%lld\n", sum[pos[n]]+ad[pos[n]]);
	return 0;
}