#include<algorithm>
#include<cstdio>
#define Maxn 1000500
using namespace std;
inline int read()
{
  register int X=0;
  register char ch=0;
  while(ch<48||ch>57)ch=getchar();
  while(ch>=48&&ch<=57)X=X*10+(ch^48),ch=getchar();
  return X;
}
int n,m;
long long f[Maxn],g[Maxn],s[Maxn];
void mul(long long *s,long long *f,long long *g)
{
  for (int k=0;k<n+m-1;k++)
    for (int i=0;i<=k;i++)
      s[k]+=f[i]*g[k-i];
}
int main()
{
  scanf("%d%d",&n,&m);n++;m++;
  for (int i=0;i<n;i++)f[i]=read();
  for (int i=0;i<m;i++)g[i]=read();
  mul(s,f,g);
  for (int i=0;i<n+m-1;i++)printf("%lld ",s[i]);
  return 0;
}

#include <iostream>
using namespace std;
struct CP
{
  CP (double xx=0,double yy=0){x=xx,y=yy;}
  double x,y;
  CP operator + (CP const &B) const
  {return CP(x+B.x,y+B.y);}
  CP operator - (CP const &B) const
  {return CP(x-B.x,y-B.y);}
  CP operator * (CP const &B) const
  {return CP(x*B.x-y*B.y,x*B.y+y*B.x);}
  CP operator / (CP const &B) const {
    double t=B.x*B.x+B.y*B.y;
    return CP((x*B.x+y*B.y)/t,
              (y*B.x-x*B.y)/t);
  } 
}a,b;
int main()
{
  cin>>a.x>>a.y>>b.x>>b.y;
  CP c;
  c=a+b;
  cout<<'('<<c.x<<','<<c.y<<")\n";
  c=a-b;
  cout<<'('<<c.x<<','<<c.y<<")\n";
  c=a*b;
  cout<<'('<<c.x<<','<<c.y<<")\n";
  c=a/b;
  cout<<'('<<c.x<<','<<c.y<<")\n";
  return 0;
}

#include<cstdio>
#include<cmath>
#define Maxn 1000500
//用这句话能得到得到精确的π,可以当做结论来记
const double Pi=acos(-1);
using namespace std;
struct CP
{
  CP (double xx=0,double yy=0){x=xx,y=yy;}
  double x,y;
  CP operator + (CP const &B) const
  {return CP(x+B.x,y+B.y);}
  CP operator - (CP const &B) const
  {return CP(x-B.x,y-B.y);}
  CP operator * (CP const &B) const
  {return CP(x*B.x-y*B.y,x*B.y+y*B.x);}
  //除法没用
}w[Maxn];
//w长得是不是很像ω?
int n;
int main()
{
  scanf("%d",&n);
  CP sav(cos(2*Pi/n),sin(2*Pi/n)),buf(1,0);
  for (int i=0;i<n;i++){
    w[i]=buf;
	  buf=buf*sav;
  }
  for (int i=0;i<n;i++)
	  printf("w[%d][n]=(%.4lf,%.4lf)\n",i,w[i].x,w[i].y);
  //由于精度问题会出现-0.0000的情况,将就看吧
  return 0;
}

#include <cstdio>
#include <cmath>
#define Maxn 1350000
using namespace std;
const double Pi=acos(-1);
int n,m;
struct CP
{
  CP (double xx=0,double yy=0){x=xx,y=yy;}
  double x,y;
  CP operator + (CP const &B) const
  {return CP(x+B.x,y+B.y);}
  CP operator - (CP const &B) const
  {return CP(x-B.x,y-B.y);}
  CP operator * (CP const &B) const
  {return CP(x*B.x-y*B.y,x*B.y+y*B.x);}
  //除法没用
}f[Maxn<<1],sav[Maxn<<1];
void dft(CP *f,int len)
{
  if (len==1)return ;//边界
  //指针的使用比较巧妙 
  CP *fl=f,*fr=f+len/2;
  for (int k=0;k<len;k++)sav[k]=f[k];
  for (int k=0;k<len/2;k++)//分奇偶打乱
    {fl[k]=sav[k<<1];fr[k]=sav[k<<1|1];}
  dft(fl,len/2);
  dft(fr,len/2);//处理子问题
  //由于每次使用的单位根次数不同(len次单位根),所以要重新求。
  CP tG(cos(2*Pi/len),sin(2*Pi/len)),buf(1,0);
  for (int k=0;k<len/2;k++){
    //这里buf = (len次单位根的第k个) 
    sav[k]=fl[k]+buf*fr[k];//(1)
    sav[k+len/2]=fl[k]-buf*fr[k];//(2)
    //这两条语句具体见Tips的式子
    buf=buf*tG;//得到下一个单位根。
  }for (int k=0;k<len;k++)f[k]=sav[k];
}
int main()
{
  scanf("%d",&n);
  for (int i=0;i<n;i++)scanf("%lf",&f[i].x);
  //一开始都是实数,虚部为0
  for(m=1;m<n;m<<=1);
  //把长度补到2的幂,不必担心高次项的系数,因为默认为0
  dft(f,m);
  for(int i=0;i<m;++i)
    printf("(%.4f,%.4f)\n",f[i].x,f[i].y);
  return 0;
}

#include <cstdio>
#include <cmath>
#define Maxn 1350000
using namespace std;
const double Pi=acos(-1);
int n,m;
struct CP
{
  CP (double xx=0,double yy=0){x=xx,y=yy;}
  double x,y;
  CP operator + (CP const &B) const
  {return CP(x+B.x,y+B.y);}
  CP operator - (CP const &B) const
  {return CP(x-B.x,y-B.y);}
  CP operator * (CP const &B) const
  {return CP(x*B.x-y*B.y,x*B.y+y*B.x);}
  //除法没用
}f[Maxn<<1],p[Maxn<<1],sav[Maxn<<1];
void dft(CP *f,int len)
{
  if (len==1)return ;//边界
  //指针的使用比较巧妙 
  CP *fl=f,*fr=f+len/2;
  for (int k=0;k<len;k++)sav[k]=f[k];
  for (int k=0;k<len/2;k++)//分奇偶打乱
    {fl[k]=sav[k<<1];fr[k]=sav[k<<1|1];}
  dft(fl,len/2);
  dft(fr,len/2);//处理子问题
  //由于每次使用的单位根次数不同(len次单位根),所以要重新求。
  CP tG(cos(2*Pi/len),sin(2*Pi/len)),buf(1,0);
  for (int k=0;k<len/2;k++){
    //这里buf = (len次单位根的第k个) 
    sav[k]=fl[k]+buf*fr[k];//(1)
    sav[k+len/2]=fl[k]-buf*fr[k];//(2)
    //这两条语句具体见Tips的式子
    buf=buf*tG;//得到下一个单位根。
  }for (int k=0;k<len;k++)f[k]=sav[k];
}
void idft(CP *f,int len)
{
  if (len==1)return ;//边界
  //指针的使用比较巧妙 
  CP *fl=f,*fr=f+len/2;
  for (int k=0;k<len;k++)sav[k]=f[k];
  for (int k=0;k<len/2;k++)//分奇偶打乱
    {fl[k]=sav[k<<1];fr[k]=sav[k<<1|1];}
  idft(fl,len/2);
  idft(fr,len/2);//处理子问题
  CP tG(cos(2*Pi/len),  -  sin(2*Pi/len)),buf(1,0);
               //注意这 ↑ 个负号! 
  for (int k=0;k<len/2;k++){
    //这里buf = (len次反单位根的第k个) 
    sav[k]=fl[k]+buf*fr[k];
    sav[k+len/2]=fl[k]-buf*fr[k];
    buf=buf*tG;//得到下一个反单位根。
  }for (int k=0;k<len;k++)f[k]=sav[k];
}
int main()
{
  scanf("%d%d",&n,&m);
  for (int i=0;i<=n;i++)scanf("%lf",&f[i].x);
  for (int i=0;i<=m;i++)scanf("%lf",&p[i].x);
  for(m+=n,n=1;n<=m;n<<=1);//把长度补到2的幂
  dft(f,n);dft(p,n);//DFT
  for(int i=0;i<n;++i)f[i]=f[i]*p[i];//点值直接乘
  idft(f,n);//IDFT
  for(int i=0;i<=m;++i)printf("%d ",(int)(f[i].x/n+0.49));
  //注意结果除以n
  return 0;
}