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 <cstring>
#include <algorithm>
#include <vector>
using namespace std;
typedef long long ll;
#define GG puts("FUCK")
#define N 100050
int head[N],to[N],nxt[N],fa[N],a[N],b[N],c[N],n,m,vis[N],id[N],cnt;
vector<int>v[N];
ll f[N],s[N];
int find(int x) {return fa[x]==x?x:fa[x]=find(fa[x]);}
bool cmp(int x,int y) {return c[x]<c[y];}
inline void add(int u,int v) {
	to[++cnt]=v; nxt[cnt]=head[u]; head[u]=cnt;
}
void dfs(int x) {
	int i; s[x]=b[x];
	for(i=head[x];i;i=nxt[i]) {
		dfs(to[i]); s[x]+=s[to[i]];
	}
	f[x]=s[x]+c[x];
	for(i=head[x];i;i=nxt[i]) {
		f[x]=min(f[x],s[x]-s[to[i]]+max(f[to[i]],1ll*c[x]));
	}
}
int main() {
	scanf("%d%d",&n,&m);
	int i,x,y,j;
	for(i=1;i<=n;i++) {
		scanf("%d%d",&a[i],&b[i]); c[i]=max(a[i]-b[i],0); fa[i]=id[i]=i;
	}
	for(i=1;i<=m;i++) {
		scanf("%d%d",&x,&y); v[x].push_back(y); v[y].push_back(x);
	}
	sort(id+1,id+n+1,cmp);
	for(i=1;i<=n;i++) {
		x=id[i]; int lim=v[x].size(); vis[x]=1;
		for(j=0;j<lim;j++) {
			y=v[x][j]; if(!vis[y]) continue;
			y=find(y);
			if(x!=y) {
				fa[y]=x; add(x,y);
			}
		}
	}
	dfs(id[n]);
	printf("%lld\n",f[id[n]]);
}