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;
typedef long long ll;
const int N=1e5+5;
int n,m,f[N],id[N],a[N],b[N],c[N],tot,head[N],nex[N<<1],to[N<<1];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
ll dp[N],s[N];
vector<int>e[N];
bool vis[N];
int getf(int x){return f[x]==x?x:f[x]=getf(f[x]);}
bool cmp(int x,int y){return c[x]<c[y];}
void dfs(int u)
{
    s[u]=b[u];
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];
        dfs(v);
        s[u]+=s[v];
    }
    dp[u]=s[u]+c[u];
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];
        dp[u]=min(dp[u],s[u]-s[v]+max(dp[v],(ll)c[u]));
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    ll ans=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&a[i],&b[i]);
        c[i]=max(a[i]-b[i],0);
        f[i]=id[i]=i;
    }
    for(int i=1;i<=m;i++)
    {
        int u,v;scanf("%d%d",&u,&v);
        e[u].push_back(v);
        e[v].push_back(u);
    }
    sort(id+1,id+1+n,cmp);
    for(int i=1;i<=n;i++)
    {
        int x=id[i];vis[x]=true;
        for(int j=0;j<e[x].size();j++)
        {
            int y=e[x][j];if(!vis[y]) continue;
            y=getf(y);
            if(x!=y)
            {
                f[y]=x;
                add(x,y);
            }
        }
    }
    dfs(id[n]);
    printf("%lld\n",dp[id[n]]);
}
