// Sometimes, the questions are complicated - and the answers are simple. //
#include<iostream>
#include <bits/stdc++.h>
#define ll long long
#define ld  long double
#define IO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int N=5e5+9, mod=1e9+7;;
int t, n, cnt[61], val[61];
ll a[N];
int main() {
    IO

    for(ll j=0 ; j<60 ; j++)
        val[j]=(1LL<<j)%mod;

    cin>>t;
    while(t--) {
        cin>>n;

        ll ans=0, s=0;
        memset(cnt,0,sizeof cnt);

        for(int i=1; i<=n; i++) {
            cin>>a[i];
            s=(s+a[i]%mod);
            for(ll j=0 ; j<60 ; j++)
                if(((1LL<<j)&a[i])!=0)
                    cnt[j]++;
        }

        s%=mod;

        for(int i=1; i<=n; i++) {
            ll sor=0;
            ll sum2=0;
            for(int j=0; j<60; j++) {
                if(((1LL<<j)&a[i])!=0) {

                    ans=(ans + 1LL * val[j] * s % mod * n)%mod;
                    sum2=(sum2 + 1LL * val[j] * n % mod * n)%mod;
                    sor=(sor + 1LL* val[j] *n)%mod;

                } else {
                    ans=(ans + 1LL * val[j] * s % mod  * cnt[j])%mod;
                    sum2=(sum2 + 1LL * val[j] * n % mod * cnt[j])%mod;
                    sor=(sor + 1LL* val[j] *cnt[j])%mod;
                }
            }
            sor=sor*sor%mod;
            sum2=sum2 *(a[i]%mod) %mod;
            ans=(ans-sor)%mod;
            ans=(ans+sum2)%mod;
        }
        cout<<(ans+mod)%mod<<"\n";
    }

}
