The Boss on Mars

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2327 Accepted Submission(s): 718

Problem Description

On Mars, there is a huge company called ACM (A huge Company on Mars), and it’s owned by a younger boss.

Due to no moons around Mars, the employees can only get the salaries per-year. There are n employees in ACM, and it’s time for them to get salaries from their boss. All employees are numbered from 1 to n. With the unknown reasons, if the employee’s work number is k, he can get k^4 Mars dollars this year. So the employees working for the ACM are very rich.

Because the number of employees is so large that the boss of ACM must distribute too much money, he wants to fire the people whose work number is co-prime with n next year. Now the boss wants to know how much he will save after the dismissal.

Input

The first line contains an integer T indicating the number of test cases. (1 ≤ T ≤ 1000) Each test case, there is only one integer n, indicating the number of employees in ACM. (1 ≤ n ≤ 10^8)

Output

For each test case, output an integer indicating the money the boss can save. Because the answer is so large, please module the answer with 1,000,000,007.

Sample Input

2 4 5

Sample Output

82 354

Hint

Case1: sum=1+3*3*3*3=82 Case2: sum=1+2*2*2*2+3*3*3*3+4*4*4*4=354

http://acm.hdu.edu.cn/showproblem.php?pid=4059

///这题主要是公式问题，然后稍微对容斥原理计算个数，改为mult的特殊利用即可；
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
///1~n这n个数的4次方和：n*(n+1)*(6*n*n*n+9*n*n+n-1)/30； 这里百度的公式；
typedef long long ll;
const ll N = 1e6+5;
const ll mod = 1e9+7;
ll n, ans;
 
ll get4mult(ll mult)
{
    return mult*mult%mod*mult%mod*mult%mod;
}
ll extgcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0){
        x = 1, y = 0; return a;
    }
    ll d = extgcd(b,a%b,x,y);
    ll t = x;
    x = y;
    y = t-a/b*y;
    return d;
}
ll inverse(ll t)
{
    ll x, y;
    ll d = extgcd(t,mod,x,y);
    return (x%mod+mod)%mod;
}
ll calu(ll n)
{
    return n*(n+1)%mod*(6*n*n%mod*n%mod+9*n*n%mod+n-1)%mod*inverse(30)%mod;
}
ll rongchi(ll n)
{
    ll m = n;
    vector<ll> v;
    for(ll i = 2; i*i <= m; i++){
        if(m%i==0){
            v.push_back(i);
            while(m%i==0) m/=i;
        }
    }
    if(m>1) v.push_back(m);
    ll len = v.size();
    for(ll i = 1; i < (1<<len); i++){
        ll mult = 1, ones = 0;
        for(ll j = 0; j < len; j++){
            if(i&(1<<j)){
                ones++;
                mult = mult*v[j];
            }
        }
        if(ones%2) {/// 变成了 -  因为假设是原来的容斥，那么cnt指的是不互质的个数，是要被剪掉的；
                ///而本题：也是要减掉，但是不是计算个数，而是对每一个过程有作用，减去：减变为了加，加变为了减；
            ans = ans-get4mult(mult)*calu(n/mult);
            ans = (ans%mod+mod)%mod;
        }else/// 变成了 +
        {
            ans = ans+get4mult(mult)*calu(n/mult);
            ans = (ans%mod+mod)%mod;
        }
    }
    v.clear();
    return ans;
}
int main()
{
    ll T;
    cin>>T;
    while(T--)
    {
        cin>>n;
        ans = calu(n);
        printf("%lld
",rongchi(n));
    }
    return 0;
}