//By TheOneYouWant
#pragma GCC optimize ("-O2")
#include <bits/stdc++.h>
using namespace std;
#define fastio ios_base::sync_with_stdio(0);cin.tie(0)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define memreset(a) memset(a,0,sizeof(a))
#define testcase(t) int t;cin>>t;while(t--)
#define forstl(i,v) for(auto &i: v)
#define forn(i,e) for(int i=0;i<e;++i)
#define forsn(i,s,e) for(int i=s;i<e;++i)
#define rforn(i,s) for(int i=s;i>=0;--i)
#define rforsn(i,s,e) for(int i=s;i>=e;--i)
#define bitcount(a) __builtin_popcount(a) // set bits (add ll)
#define ln '\n'
#define getcurrtime() cerr<<"Time = "<<((double)clock()/CLOCKS_PER_SEC)<<endl
#define dbgarr(v,s,e) cerr<<#v<<" = "; forsn(i,s,e) cerr<<v[i]<<", "; cerr<<endl
#define inputfile freopen("input.txt", "r", stdin)
#define outputfile freopen("output.txt", "w", stdout)
template<typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
	cerr << *it << " = " << a << "\t"; err(++it, args...);
}
template<typename T1,typename T2>
ostream& operator <<(ostream& c,pair<T1,T2> &v){
	c<<"("<<v.fi<<","<<v.se<<")"; return c;
}
template <template <class...> class TT, class ...T>
ostream& operator<<(ostream& out,TT<T...>& c){
    out<<"{ ";
    forstl(x,c) out<<x<<" ";
    out<<"}"; return out;
}
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll,ll> p64;
typedef pair<int,int> p32;
typedef pair<int,p32> p96;
typedef vector<ll> v64;
typedef vector<int> v32; 
typedef vector<v32> vv32;
typedef vector<v64> vv64;
typedef vector<p32> vp32;
typedef vector<p64> vp64;
typedef vector<vp32> vvp32;
typedef map<int,int> m32;
const int LIM=2e5+5,MOD=1e9+7;
const ld EPS = 1e-9;


mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());


void solve() {
	ll n;
	cin >> n;
	ll ans = 0;
	//count small b
	ll b = 1;

	ll mxb = 1;
	for(;mxb*mxb <= n; mxb++);

	for(;b*b <= n; b++) {
		ll a = 1;
		for(;a*a <= n/b; a++) {
			ans += n/(a*b+1);
		}

		for(ll k = 1; k <= n/(a*b+1); k++) {
			ll maxval = (n-k)/(b*k);
			ll minval = (n-k-1)/(b*(k+1)) + 1;
			if(k == n/(a*b+1))
				minval = max(minval, a);
			if(maxval >= minval)
				ans += k*(maxval - minval + 1);
			ll cut = max(mxb, minval);
			if(maxval >= cut)
				ans += k*(maxval - cut+1);
		}
	}
	cout << ans << ln;

}

int main(){
	fastio;
	int t;
	cin >> t;
	while(t--)
		solve();
	

	return 0;
}