分别考虑每一种颜色对答案的贡献。每种颜色的贡献就是他出现的区间个数，那么可以用总区间减去不包含它的区间个数，把每一个序列里不包含它的区间个数加起来，然后不同序列用乘法原理计算即可

于是我辛辛苦苦打了两个小时交上去只剩两分……后来看了题解之后才发现我忘了考虑某个序列全都是一个数的情况……那种情况下这个数贡献为0，然而后面删除它的时候没办法消除贡献因为0没有逆元……

于是解决方法就是记录一下这个数字是否在某个序列里全都是，如果是的话不包含它的区间为0，否则的话正常计算就可以了

然后代码里是一些丧心病狂的卡常……~~抄了iscream巨巨的题解还不小心抢了他的rank1心里有点莫名其妙的羞愧→_→~~

//minamoto#include
    
     #define R register#define IT set
     
      ::iterator#define fp(i,a,b) for(R int i=a,I=b+1;i
      
       I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    R int res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}char sr[1<<21],z[20];int C=-1,Z=0;inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}void print(R int x){    if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;    while(z[++Z]=x%10+48,x/=10);    while(sr[++C]=z[Z],--Z);sr[++C]='\n';}const int N=2e5+5,P=19260817;inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}inline int calc(R int l){return 1ll*l*(l+1)/2%P;}int ksm(R int x,R int y){    R int res=1;    for(;y;y>>=1,x=1ll*x*x%P)if(y&1)res=1ll*res*x%P;    return res;}set
       
        s[N];map
        
         mp[N];int len[N],sum[N],isum[N],inv[N],sl[N],zero[N];int a[N],b[N],x[N],y[N],d[N],hang[N],lie[N],nxt[N],head[N],ans[N];int n,m,tot,lim,all,loli;inline int Inv(R int x){return x<=2e5?inv[x]:ksm(x,P-2);}void solve(int x,int y,int d){    int id=sl[x-1]+y,l=0,r=len[x]+1,col=a[id];    IT itl,itr;itl=s[col].lower_bound(id),itr=itl;    if(itl!=s[col].begin()){        --itl;if(hang[*itl]==x)l=lie[*itl];    }    ++itr;if(itr!=s[col].end()&&hang[*itr]==x)r=lie[*itr];    loli=dec(loli,zero[col]?all:dec(all,ans[col]));    if(mp[col][x]>0)ans[col]=1ll*ans[col]*Inv(mp[col][x])%P;    else --zero[col];    mp[col][x]=dec(mp[col][x],add(calc(r-y-1),calc(y-l-1)));    mp[col][x]=add(mp[col][x],calc(r-l-1));    if(mp[col][x]>0)ans[col]=1ll*ans[col]*mp[col][x]%P;    else ++zero[col];    loli=add(loli,zero[col]?all:dec(all,ans[col]));    s[col].erase(id);        l=0,r=len[x]+1,a[id]=d,col=d,s[col].insert(id);    itl=s[col].lower_bound(id),itr=itl;    if(itl!=s[col].begin()){        --itl;if(hang[*itl]==x)l=lie[*itl];    }    ++itr;if(itr!=s[col].end()&&hang[*itr]==x)r=lie[*itr];    if(r==len[x]+1&&l==0)mp[col][x]=calc(len[x]);    loli=dec(loli,zero[col]?all:dec(all,ans[col]));    if(mp[col][x]>0)ans[col]=1ll*ans[col]*Inv(mp[col][x])%P;    else --zero[col];    mp[col][x]=dec(mp[col][x],calc(r-l-1));    mp[col][x]=add(mp[col][x],add(calc(r-y-1),calc(y-l-1)));    if(mp[col][x]>0)ans[col]=1ll*ans[col]*mp[col][x]%P;    else ++zero[col];    loli=add(loli,zero[col]?all:dec(all,ans[col]));    print(loli);}int main(){//  freopen("testdata.in","r",stdin);    inv[0]=inv[1]=1;for(int i=2;i<=2e5;++i)inv[i]=((-(1ll*P/i)*inv[P%i])%P+P)%P;    n=read(),m=read(),sum[0]=isum[0]=1,tot=0;    fp(i,1,n){        len[i]=read(),sl[i]=sl[i-1]+len[i];        sum[i]=1ll*calc(len[i])*sum[i-1]%P,isum[i]=Inv(sum[i]);    }all=sum[n];    fp(i,1,n)fp(j,1,len[i])b[++lim]=a[++tot]=read(),hang[tot]=i,lie[tot]=j;    fp(i,1,m)x[i]=read(),y[i]=read(),b[++lim]=d[i]=read();    sort(b+1,b+1+lim),lim=unique(b+1,b+1+lim)-b-1;    tot=0;fp(i,1,n)fp(j,1,len[i]){        ++tot,a[tot]=lower_bound(b+1,b+1+lim,a[tot])-b;        s[a[tot]].insert(tot);    }fd(i,tot,1)nxt[i]=head[a[i]],head[a[i]]=i;    fp(i,1,m)d[i]=lower_bound(b+1,b+1+lim,d[i])-b;    fp(i,1,lim)if(head[i]){        ans[i]=sum[hang[head[i]]-1];        for(R int j=head[i];j;j=nxt[j]){            int bl=hang[j],res=0,las=sl[bl-1];            while(hang[nxt[j]]==bl)res=add(res,calc(j-las-1)),las=j,j=nxt[j];            res=add(res,calc(j-las-1)),res=add(res,calc(sl[bl]-j));            if(res)ans[i]=1ll*ans[i]*res%P,mp[i][bl]=res;            else ++zero[i];            if(!nxt[j])ans[i]=1ll*ans[i]*sum[n]%P*isum[bl]%P;            else ans[i]=1ll*ans[i]*sum[hang[nxt[j]]-1]%P*isum[bl]%P;        }loli=add(loli,zero[i]?all:dec(all,ans[i]));    }else ans[i]=sum[n];    print(loli);fp(i,1,m)solve(x[i],y[i],d[i]);return Ot(),0;}