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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
typedef long long ll;
const int N=100010;
bool w[N];
long long sz[N];
int fa[N],a[N],b[N],Mx[N],id[N];
vector <int> q[N];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
inline bool cmp(const int &x,const int &y) {return a[x]<a[y];}
inline int find(int x) {return fa[x]==x?x:fa[x]=find(fa[x]);}
int main() {
    int n,m;
    cin>>n>>m;
    for(int i=1;i<=n;i++) {
	a[i]=gi(),b[i]=gi(),id[i]=i,fa[i]=i;
	Mx[i]=a[i]=max(0,a[i]-b[i]),sz[i]=b[i];
    }
    for(int i=1;i<=m;i++) {
	int x=gi(),y=gi();
	q[x].push_back(y),q[y].push_back(x);
    }
    sort(id+1,id+1+n,cmp);
    for(int i=1;i<=n;i++) {
	int x=id[i];w[x]=1;
	for(int j=0;j<q[x].size();j++)
	    if(w[q[x][j]]) {
		int y=find(q[x][j]);
		if(x!=y) {
		    sz[x]+=sz[y],fa[y]=x;
		    Mx[x]=min(1LL*Mx[x],Mx[y]+max(0LL,a[x]-sz[y]-Mx[y]));
		}
	    }
	if(i==n) cout<<Mx[x]+sz[x];
    }
    return 0;
}