Function

#include <atcoder/modint>
#include <bits/stdc++.h>
 
using namespace std;
using namespace atcoder;
 
using mint = modint1000000007;
 
template <class T> using V = vector<T>;
#define all(x) begin(x), end(x)
 
using ll = long long;
 
ll sq(ll x) { return x * x; }
 
// BeginCodeSnip{Dirichlet Inverse}
struct Dirichlet {
	ll N;
	int s = 0, k;
	Dirichlet(ll N_) : N(N_) {
		while (sq(s + 1) <= N) ++s;
		k = 2 * s - (s == N / s);
	}
	ll pos_to_val(int i) {
		if (i < s) return i + 1;
		return N / (k - i);
	}
	int val_to_pos(ll l) {
		if (l <= s) return l - 1;
		return k - N / l;
	}
	V<mint> invert(const V<mint> &f, const V<mint> &h) {
		// return g s.t. f * g = h
		V<mint> g(k);
		mint inv_f0 = mint(1) / f.front();
		for (int i = 0; i < k; ++i) {
			mint remainder = h.at(i);
			if (i > 0) {
				ll v = pos_to_val(i);
				ll j = 1;
				for (; j * j <= v; ++j) {
					int p = val_to_pos(v / j);
					if (j > 1) remainder -= (f[j - 1] - (j == 1 ? 0 : f[j - 2])) * g[p];
					remainder -= (g[j - 1] - (j == 1 ? 0 : g[j - 2])) * f[p];
				}
				--j;
				assert(j > 0);
				remainder += f[j - 1] * g[j - 1];
			}
			g.at(i) = remainder * inv_f0;
		}
		return g;
	}
};
// EndCodeSnip
 
void solve() {
	ll N;
	cin >> N;
	Dirichlet dc(N);
	int k = dc.k;
	V<mint> f(k), h(k);
	mint i3 = mint(1) / 3;
	for (int i = 0; i < k; ++i) {
		mint v = dc.pos_to_val(i);
		f.at(i) = v;
		h.at(i) = v * (v - 1) * (v - 2) * i3;
	}
	auto ret = dc.invert(f, h);
	cout << ret.back().val() << "\n";
}
 
int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int T;
	cin >> T;
	while (T--) solve();
}