#pragma GCC optimization ("O3")

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vec = vector<ll>;
using mat = vector<vec>;
using pll = pair<ll,ll>;

#define INF (1LL<<61)
//#define MOD 1000000007LL 
#define MOD 998244353LL
#define EPS (1e-10)

#define PR(x) cout << (x) << endl
#define PS(x) cout << (x) << " "
#define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i))
#define FORE(i,v) for(auto (i):v)
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((ll)(x).size())
#define REV(x) reverse(ALL((x)))
#define ASC(x) sort(ALL((x)))
#define DESC(x) {ASC((x)); REV((x));}
#define BIT(s,i) (((s)>>(i))&1)
#define pb push_back
#define fi first
#define se second

template<class T> inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;}
template<class T> inline int chmax(T& a, T b) {if(a<b) {a=b; return 1;} return 0;}
class mint {
public:
    ll x;
    mint(ll x=0) : x((x%MOD+MOD)%MOD) {}
    mint operator-() const {return mint(-x);}
    mint& operator+=(const mint& a) {if((x+=a.x)>=MOD) x-=MOD; return *this;}
    mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;}
    mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;}
    mint operator+(const mint& a) const {mint b(*this); return b+=a;}
    mint operator-(const mint& a) const {mint b(*this); return b-=a;}
    mint operator*(const mint& a) const {mint b(*this); return b*=a;}
    mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;}
    mint inv() const {return pow(MOD-2);}
    mint& operator/=(const mint& a) {return *this*=a.inv();}
    mint operator/(const mint& a) const {mint b(*this); return b/=a;}
};
istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;}
ostream &operator<<(ostream& os, const mint& a) {return os<<a.x;}

vec factorization(ll n)
{
    vec p;
    ll t = n;
    for(ll i=2; i*i<=t; ++i) {
        if(t%i) continue;
        while(t%i == 0) t /= i;
        p.pb(i);
    }
    if(t != 1) p.pb(t);
    return p;
}

ll modpow(ll a, ll b, ll m)
{
    if(b == 0) return 1;
    ll t = modpow(a, b>>1, m);
    return ((b&1)?a*t%m:t)*t%m;
}

ll root(ll n)
{
    vec P = factorization(n-1);
    while(1) {
        ll r = n==2?1:rand()%(n-2)+2;
        bool f = true;
        FORE(p,P) {
            if(modpow(r, (n-1)/p, n) == 1) {
                f = false;
                break;
            }
        }
        if(f) return r;
    }
    return -1;
}

void solve()
{
    ll V, X;
    cin >> V >> X;
    
    ll P = V*X+1;
    ll R = root(P);
    vec A(X);
    REP(i,0,X) A[i] = modpow(R, i*V, P);
    ASC(A);
    FORE(a,A) PS(a);
    PR("");
}

int main()
{
    ll T;
    cin >> T;
    REP(i,0,T) solve();

    return 0;
}

/*



*/