思路：求出p的原根，那麼乘法就變成了加法，FFT跑一下，下標在稍微轉換一下就好了。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(998244353)
#define pb push_back
#define eps 1e-6
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
ll n,p,g,id[200008],a[200008],m,ans[500008];
bool as(ll x)
{
    ll tmp=1;
    FOR(i,1,p-2)
    {
        tmp=tmp*x%p;
        if(tmp==1) return 0;
    }
    return 1;
}
struct CP{
    double x,y;
    CP(){} CP(double a,double b):x(a),y(b){}
    CP operator+(const CP&r) const{return CP(x+r.x,y+r.y);}
    CP operator-(const CP&r) const{return CP(x-r.x,y-r.y);}
    CP operator*(const CP&r) const{return CP(x*r.x-y*r.y,x*r.y+y*r.x);}
}b[500008],t;
inline void Swap(CP&a,CP&b) {t=a;a=b;b=t;}
inline void fft(CP*a,int f,int n)
{
    int i,j,k;
    for(i=j=0;i<n;i++)
    {
        if(i>j) Swap(a[i],a[j]);
        for(k=n>>1;(j^=k)<k;k>>=1);
    }
    for(int i=1;i<n;i<<=1)
    {
        CP wn(cos(pi/i),f*sin(pi/i));
        for(int j=0;j<n;j+=i<<1)
        {
            CP w(1,0);
            for(int k=0;k<i;k++,w=w*wn)
            {
                CP x=a[j+k],y=w*a[i+j+k];
                a[j+k]=x+y;a[i+j+k]=x-y;
            }
        }
    }
    if(f==-1) FOR(i,0,n) a[i].x/=n;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n>>p;
    for(ll i=2;;i++) if(as(i)) {g=i;break;}
    ll tmp=1,qw=0,er;
    for(int i=1;i<p-1;i++)
    {
        tmp=tmp*g%p;
        id[tmp]=i;
    }
    FOR(i,1,n)
    {
        sl(a[i]);
        if(!(a[i]%p)) qw++;
        else b[id[a[i]%p]].x+=1.0;
    }
    er=m=p-1;
    for(m=er+m,er=1;er<=m;er<<=1);
    fft(b,1,er);
    FOR(i,0,er) b[i]=b[i]*b[i];
    fft(b,-1,er);
    for(int i=0;i<2*p;i++) ans[i%(p-1)]+=(ll)(b[i].x+0.2);
    //cout<<ans[id[1]]<<endl;
    for(int i=1;i<=n;i++)
    {
        if(a[i]>=p) puts("0");
        else if (a[i]==0) printf("%lld\n",2*qw*n-qw*qw);
        else printf("%lld\n",ans[id[a[i]]]);
    }
    return 0;
}