注意：n边形里共有n-3条边
最优步数=不与n相连的边数，关键是方案数．
按照处理顺序可以转化为树形结构即二叉树森林，转移方案数用组合数即可
关键是快速处理修改．
1.最优解减少一步，即删掉某棵二叉树的根，合并它的两个儿子．
2.相当于在splay中把它rotate一下，而且不知道为什么它还一定是左儿子
#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<cassert>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define Debug(x) cout<<#x<<"="<<x<<endl
using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
const int INF=1e9+7;
inline LL read(){
register LL x=0,f=1;register char c=getchar();
while(c<48||c>57){if(c=='-')f=-1;c=getchar();}
while(c>=48&&c<=57)x=(x<<3)+(x<<1)+(c&15),c=getchar();
return f*x;
}
const int N=100005;
const int mod=1e9+7;
int size[N],ls[N],rs[N],fa[N],fac[N],ifac[N],inv[N];
int W,n,m,sum,ans=1,Ecnt;
vector <int> E[N];
map <pii,int> M;
inline int add(int x,int y){x+=y;return x>=mod?x-mod:x;}
inline int mul(LL x,int y){x*=y;return x>=mod?x%mod:x;}
inline int cal(int x,int y){return mul(fac[x+y],mul(ifac[x],ifac[y]));}
inline int ical(int x,int y){return mul(ifac[x+y],mul(fac[x],fac[y]));}
inline void dfs(int &x,int l,int r,int pre){
if(l+1==r) return;
x=++Ecnt;///
M[pii(l,r)]=x,fa[x]=pre;
int mid=*upper_bound(E[r].begin(),E[r].end(),l);
dfs(ls[x],l,mid,x);
dfs(rs[x],mid,r,x);
size[x]=size[ls[x]]+size[rs[x]]+1;
ans=mul(ans,cal(size[ls[x]],size[rs[x]]));
}
inline void answer(int x,int y){
if(W==0) printf("%d
",x);
else printf("%d %d
",x,y);
}
int main(){
W=read(),n=read();
inv[0]=inv[1]=fac[0]=fac[1]=ifac[0]=ifac[1]=1;
for(int i=2;i<=n;i++) inv[i]=mul(inv[mod%i],mod-mod/i);///
for(int i=2;i<=n;i++) fac[i]=mul(fac[i-1],i),ifac[i]=mul(ifac[i-1],inv[i]);
for(int i=1;i<=n-3;i++){
int x=read(),y=read();
E[x].push_back(y);E[y].push_back(x);
}
for(int i=1;i<n;i++) E[i].push_back(i+1),E[i+1].push_back(i);
E[1].push_back(n),E[n].push_back(1);
for(int i=1;i<=n;i++) sort(E[i].begin(),E[i].end());
for(int i=0;i<(E[n].size()-1);i++){
int tmp=0;
dfs(tmp,E[n][i],E[n][i+1],0);
ans=mul(ans,cal(sum,size[tmp]));
sum+=size[tmp];
}
answer(sum,ans);
m=read();
for(int i=1;i<=m;i++){
int a=read(),b=read(),x=M[pii(a,b)];
int qans=ans,qsum=sum;//-(a==n||b==n);
if(!fa[x]){
//assert(qsum==sum-1);
qsum--;
qans=mul(qans,ical(size[ls[x]],size[rs[x]]));
qans=mul(qans,ical(sum-size[x],size[x]));
qans=mul(qans,cal(sum-size[x],size[ls[x]]));
qans=mul(qans,cal(sum-size[x]+size[ls[x]],size[rs[x]]));
}
else{
int y=fa[x],k=(rs[y]==x);
qans=mul(qans,ical(size[ls[x]],size[rs[x]]));
qans=mul(qans,ical(size[ls[y]],size[rs[y]]));
if(k==0){
qans=mul(qans,cal(size[rs[x]],size[rs[y]]));
qans=mul(qans,cal(size[ls[x]],size[rs[x]]+size[rs[y]]+1));
}
if(k==1){
assert(false);
qans=mul(qans,cal(size[ls[y]],size[ls[x]]));
qans=mul(qans,cal(size[rs[x]],size[ls[y]]+size[ls[x]]+1));
}
}
answer(qsum,qans);
}
}