#include <bits/stdc++.h>

using namespace std;

namespace jumpmelon
{
    typedef long long big;

    const int B = 1000000, PB = 1000000, P = 1000000007, INV_2 = (P + 1) >> 1;
    bool IsPrime[B + 1];
    int pc, Prime[B + 1], Mu[B + 1], Sum[B + 1], Pre[PB + 1];

    void init()
    {
        memset(IsPrime, 1, sizeof(IsPrime));
        Mu[1] = 1;
        for (int i = 2; i <= B; i++)
        {
            if (IsPrime[i])
            {
                Prime[pc++] = i;
                Mu[i] = P - 1;
            }
            for (int j = 0; j < pc; j++)
            {
                int t = i * Prime[j];
                if (t > B)
                    break;
                IsPrime[t] = 0;
                if (i % Prime[j])
                    Mu[t] = (P - Mu[i]) % P;
                else
                {
                    Mu[t] = 0;
                    break;
                }
            }
        }
        for (int i = 1; i <= B; i++)
            Sum[i] = int((Sum[i - 1] + (big)Mu[i] * i) % P);
        for (int a = 1; a <= PB; a++)
            for (int b = 1; a * b <= PB; b++)
                Pre[a * b] = (Pre[a * b] + a) % P;
        for (int i = 2; i <= PB; i++)
            Pre[i] = (Pre[i - 1] + Pre[i]) % P;
    }

    big msqrt(big x)
    {
        big t = (big)sqrt(x);
        while ((t + 1) * (t + 1) <= x)
            t++;
        while (t * t > x)
            t--;
        return t;
    }

    inline int sum(big x) { return int(x * (x + 1) / 2 % P); }

    int calc(big n)
    {
        if (n <= PB)
            return Pre[n];
        int ans = 0;
        for (big l = 1, r; l <= n; l = r + 1)
        {
            r = n / (n / l);
            ans = int((ans + big(r - l + 1) * sum(n / l % P)) % P);
        }
        return ans;
    }

    void work()
    {
        init();
        int kase;
        scanf("%d", &kase);
        while (kase--)
        {
            big n;
            int ans = 0;
            scanf("%lld", &n);
            for (big l = 1, r; l * l <= n; l = r + 1)
            {
                big t = n / (l * l);
                r = msqrt(n / t);
                ans = int((ans + big(Sum[r] + P - Sum[l - 1]) * calc(t)) % P);
            }
            printf("%d\n", ans);
        }
    }
}

int main()
{
    jumpmelon::work();
    return 0;
}