这题纯暴力竟然能过……

1 //minamoto 2 #include
     
       3 #include
      
        4 #define mul(a,b) (1ll*a*b%P) 5 using namespace std; 6 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++) 7 char buf[1<<21],*p1=buf,*p2=buf; 8 int read(){ 9     #define num ch-'0'10     char ch;bool flag=0;int res;11     while(!isdigit(ch=getc()))12     (ch=='-')&&(flag=true);13     for(res=num;isdigit(ch=getc());res=res*10+num);14     (flag)&&(res=-res);15     #undef num16     return res;17 }18 char sr[1<<21],z[20];int C=-1,Z;19 inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}20 void print(int x){21     if(C>1<<20)Ot();if(x<0)sr[++C]=45,x=-x;22     while(z[++Z]=x%10+48,x/=10);23     while(sr[++C]=z[Z],--Z);sr[++C]='\n';24 }25 const int N=2e5+5,P=1e9+7;26 int n,m,a[N],b[N],c[N],ans=1;27 int ksm(int a,int b){28     int res=1;29     while(b){30         if(b&1) res=mul(res,a);31         a=mul(a,a),b>>=1;32     }33     return res;34 }35 int main(){36 //    freopen("testdata.in","r",stdin);37     n=read(),m=read();38     for(int i=1;i<=n;++i)39     a[i]=read(),c[i]=max(c[i-1],a[i]);40     for(int i=1;i<=n;++i)41     b[i]=read(),ans=mul(min(b[i],c[i]),ans);42     while(m--){43         int op=read(),x=read(),y=read();44         if(op==1){45             if(b[x]