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 <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int N=1e5+5;
int n,m,ID[N],fa[N],vis[N];
ll A[N],B[N],C[N],dp[N],sum[N];
vector <int> G[N];

template <typename _Tp> inline void IN(_Tp&x) {
	char ch;bool flag=0;x=0;
	while(ch=getchar(),!isdigit(ch)) if(ch=='-') flag=1;
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	if(flag) x=-x;
}
int find(int x) {return x==fa[x]?fa[x]:fa[x]=find(fa[x]);}
bool cmp(int x,int y) {return C[x]<C[y];}
inline void addedge(int u=0,int v=0) {
	IN(u),IN(v),G[u].push_back(v),G[v].push_back(u);
}

int main() {
	IN(n),IN(m);
	for(int i=1;i<=n;++i) fa[i]=ID[i]=i;
	for(int i=1;i<=n;++i) IN(A[i]),IN(B[i]),C[i]=max(A[i]-B[i],(ll)0);
	sort(ID+1,ID+1+n,cmp);
	for(int i=1;i<=m;++i) addedge();
	int u;
	for(int i=1;i<=n;++i) {
		vis[u=ID[i]]=true,sum[u]=B[u],dp[u]=C[u];
		for(auto tmp:G[u]) if(vis[tmp]) {
			int v=find(tmp);
			if(v==u) continue;
			fa[v]=u,sum[u]+=sum[v],dp[u]=min(dp[u],max(dp[v],C[u]-sum[v]));
		}
	}
	printf("%lld\n",dp[ID[n]]+sum[ID[n]]);
	return 0;
}