#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
using namespace std;
const int N = 1e7 + 5;
bool prime[N];
ll pref[N], cnt[N], np[N];
int t, n;
void get(ll &p, ll &q) {
	if(q == 0) {
		q = 1;
		return;
	}
	ll g = __gcd(p, q);
	p /= g;
	q /= g;
}
pair<ll, ll> sum(ll p1, ll q1, ll p2, ll q2) {
	ll q = q1 * q2;
	ll p = p1 * q2 + p2 * q1;
	get(p, q);
	return {p, q};
}
int main() {
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    memset(prime, true, sizeof prime);
    for(int i = 2;i < N;i++) {
		pref[i] = pref[i - 1];
		if(!prime[i])
			continue;
		pref[i] += sqrt(i) - 1;;
		for(int j = i + i;j < N;j += i) {
			if(prime[j] == true)
				cnt[j] = i - 1;
			prime[j] = false;
		}
	}
	for(int i = 2;i < N;i++) {
		np[i] = prime[i] + np[i - 1];
		cnt[i] += cnt[i - 1];
	}
    cin >> t;
    while(t--) {
		cin >> n;
		ll p = cnt[n] + pref[n], q = np[n];
		get(p, q);
		cout << p << "/" << q << '\n';
	}
    return 0;
}
//E(x) = E(x) * p + p
//E(x) * (1 - p) = p
//E(x) = p / (1 - P)