hdu 6304 多校第一场 Chiaki Sequence Revisited_chiaki is interested in an infinite sequencea1,a2,-优快云博客

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//#include <bits/stdc++.h>
#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#include<stack>
#include<map>
#include<set>
#include<stdlib.h>
#include<time.h>
#include <iomanip>
#define lowbit(x) (x&(-x))
#define inf  0x7fffffff
#define linf 0x7fffffffffffffff
#define mem(x,y) memset(x,y,sizeof(x))
#define fup(i,x,y) for(int i=(x);i<=(y);i++)
#define fdn(i,x,y) for(int i=(x);i>=(y);i--)
#define sp(x) setprecision(x)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define sld(n) scanf("%lld",&n)
#define sldd(n,m) scanf("%lld%lld",&n,&m)
#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)
#define sf(n) scanf("%lf",&n)
#define sff(n,m) scanf("%lf%lf",&n,&m)
#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)
#define sc(n) scanf("%s",n)
#define pf(x) printf("%d\n",x)
#define pfl(x) printf("%lld\n",x)
#define pff(x) printf("%lf\n",x)
#define debug printf("!!\n");
#define N 100005
#define M 4000009
#define pi acos(-1)
#define eps 1e-2
//cout.setf(ios::fixed);
//freopen("out.txt","w",stdout);// freopen("in.txt","r",stdin);
using namespace std;
typedef long long  ll;
typedef double db;
const int mod=1e9+7;
ll p[105];
ll quick(ll a,ll b)
{
    ll ans=1;
    while(b)
    {
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ans;
}
ll inv,n;
ll ok(ll x)
{
    ll sum=0;
    fup(i,0,62)
    {
        if(p[i]>x) break;
        sum+=x/p[i];
        if(sum>n-1) return 2;
    }
    if(sum==n-1) return 1;
    return 0;
}
void Solve()
{
    sld(n);
    if(n==1)
    {
        puts("1");
        return ;
    }
    ll l=max(1ll,n/2-30),r=min(n,n/2+30),mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        int f=ok(mid);
        if(f==2) r=mid-1;
        else if(f) break;
        else l=mid+1;
    }
    if(!ok(mid)) mid++;
    ll now=mid-1;
    ll ans=0,sum=0;
    for(int i=0;i<=62;i++)
    {
        if(p[i]>now) break;
        ll tp=now/p[i];
        tp%=mod;
        ll res=(((p[i]%mod*(tp+1))%mod*tp)%mod*inv)%mod;
        ans=(ans+res)%mod;
        sum+=tp;
    }
    ll last=((n-1-sum)%mod*mid)%mod;
    ans=(ans+last)%mod;
    pfl((1+ans)%mod);
}

int main()
{
    p[0]=1;
    fup(i,1,62)
        p[i]=p[i-1]*2;
    inv=(mod+1)/2;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        Solve();
    }
    return 0;
}

/*
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*/