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<cctype>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=1e5+500;
typedef long long ll;

int gti(void)
{
    char c=getchar();
    int ret=0,st=1;
    for (;!isdigit(c);c=getchar()) if (c=='-') st=-1;
    for (;isdigit(c);c=getchar()) ret=ret*10+c-'0';
    return ret*st;
}

struct edge{edge *nxt;int to;}pool[N<<1],*pt[N],*ep=pool;
void add(int a,int b)
{
    *(++ep)=(edge){pt[a],b},pt[a]=ep;
    *(++ep)=(edge){pt[b],a},pt[b]=ep;
}

struct P
{
    int a,b,id;
    bool operator<(const P &y)const
    {
        return a<y.a||(a==y.a&&id<y.id);
    }
}p[N],q[N];

int fa[N],vis[N];
int getfa(int v)
{
    if (fa[v]==v) return v;
    return fa[v]=getfa(fa[v]);
}

ll f[N],g[N];
int main(void)
{
    int n=gti(),m=gti();
    for (int i=1;i<=n;i++)
    {
        p[i].a=gti(),p[i].b=gti(),p[i].id=i;
        p[i].a=max(0,p[i].a-p[i].b),q[i]=p[i];
    }
    for (int i=1;i<=m;i++)
    {
        int u=gti(),v=gti();
        add(u,v);
    }
    sort(p+1,p+1+n);
    for (int i=1;i<=n;i++)
    {
        int v=p[i].id;
        g[v]=q[v].b,f[v]=1e18,fa[v]=v;
        for (edge *e=pt[v];e;e=e->nxt)
            if (q[e->to]<q[v]&&vis[getfa(e->to)]<i)
                g[v]+=g[fa[e->to]],vis[fa[e->to]]=i;
        for (edge *e=pt[v];e;e=e->nxt)
            if (q[e->to]<q[v])
            {
                f[v]=min(f[v],g[v]-g[fa[e->to]]+max(f[fa[e->to]],1ll*q[v].a));
                fa[fa[e->to]]=v;
            }
        if (f[v]==1e18) f[v]=q[v].a+q[v].b;
    }
    printf("%lld\n",f[p[n].id]);
    return 0;
}
