#pragma GCC optimize(2,3,"Ofast","inline","unroll-loops")
#include<bits/stdc++.h>
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
#define pii pair<ll,ll>
#define rep(i,a,b) for(ll i=(a);i<=(b);++i)
#define per(i,a,b) for(ll i=(a);i>=(b);--i)
using namespace std;
bool Mbe;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void write(ll x){
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
}
const ll N=25,Mod=1e9+7,inv2=(Mod+1)>>1;
ll n,m,U[N*N/2+5],V[N*N/2+5],cnt[N*N],f[N*N],ans[N*N];
ll C[N*N][N*N];
int ec[(1<<N)+5];
ll pw(ll x,ll p){
    ll res=1;
    while(p){
        if(p&1)res=res*x%Mod;
        x=x*x%Mod,p>>=1;
    }
    return res;
}
void FWT(ll*a,ll n,ll sgn){
    for(ll i=1;i<(1<<n);i<<=1){
        for(ll j=0;j<(1<<n);j+=(i<<1)){
            rep(k,0,i-1){
                ll x=a[j+k],y=a[j+k+i];
                a[j+k]=(x+y)*sgn%Mod;
                a[j+k+i]=(x-y+Mod)*sgn%Mod;
            }
        }
    }
}
bool Med;
int main(){
    cerr<<fabs(&Med-&Mbe)/1048576.0<<"MB\n";
    n=read(),m=read();
    rep(i,1,m)U[i]=read()-1,V[i]=read()-1;
    rep(i,1,m)ec[(1<<U[i])|(1<<V[i])]++;
    rep(i,0,n-1){
        rep(j,0,(1<<n)-1){
            if((j>>i)&1)ec[j]+=ec[j^(1<<i)];
        }
    }
    rep(i,0,(1<<n)-1){
        ll tc=m-ec[((1<<n)-1)^i]-ec[i];
        cnt[tc]++;
    }
    //tc[i] 表示只有一端在 i 集合内的边数
    //用两端至少一端在的减去两端都在的
    //那么 fwt 完之后这个位置上是 (1-y)^tc[i]*(1+y)^(m-tc[i])
    rep(i,0,m){
        C[i][0]=1;
        rep(j,1,i)C[i][j]=(C[i-1][j]+C[i-1][j-1])%Mod;
    }
    rep(i,0,m){
        memset(f,0,sizeof(f));
        rep(j,0,m){
            rep(p,0,min(i,j)){
                ll q=j-p;
                if(q>m-i)continue;
                ans[j]=(ans[j]+1ll*cnt[i]*C[i][p]%Mod*C[m-i][q]%Mod*((p&1)?Mod-1:1ll))%Mod;
            }
        }
    }
    ll ipw=pw(inv2,n);
    rep(i,0,m)write(ans[i]*ipw%Mod),putchar(' ');
    cerr<<"\n"<<clock()*1.0/CLOCKS_PER_SEC*1000<<"ms\n";
    return 0;
}