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 <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define rep(i,a) for(int i=lst[a];i;i=nxt[i])
using namespace std;

typedef long long ll;

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

const int N=1e5+5;

int t[N<<1],nxt[N<<1],lst[N],l;
void add(int x,int y) {t[++l]=y;nxt[l]=lst[x];lst[x]=l;}

int n,m,a[N],b[N],fa[N],ord[N];
bool vis[N];
ll f[N],sum[N];

vector<int> to[N];

int get(int x) {return fa[x]?fa[x]=get(fa[x]):x;}

bool cmp(int x,int y) {return a[x]<a[y];}

void merge(int x,int y) {
	int a=get(x),b=get(y);
	if (a==b) return;
	fa[b]=a;to[x].push_back(b);
}

void dfs(int x) {
	if (to[x].empty()) {
		sum[x]=b[x];f[x]=a[x];
		return;
	}
	sum[x]=b[x];
	for(int y:to[x]) dfs(y),sum[x]+=sum[y];
	f[x]=1e18;
	for(int y:to[x]) f[x]=min(f[x],max(f[y],a[x]-sum[y]));
}

int main() {
	n=read();m=read();
	fo(i,1,n) a[i]=read(),b[i]=read(),a[i]=max(0,a[i]-b[i]);
	fo(i,1,m) {
		int x=read(),y=read();
		add(x,y);add(y,x);
	}
	fo(i,1,n) ord[i]=i;
	sort(ord+1,ord+n+1,cmp);
	fo(i,1,n) {
		int x=ord[i];vis[x]=1;
		rep(j,x) {
			int y=t[j];
			if (!vis[y]) continue;
			merge(x,y);
		}
	}
	dfs(ord[n]);
	printf("%lld\n",f[ord[n]]+sum[ord[n]]);
	return 0;
}