#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
// #include <atcoder/all>
// using namespace atcoder;
using namespace std;

using ll = long long;
using pll = pair<ll, ll>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vpll = vector<pll>;
using vvpll = vector<vpll>;
#define rep(i, l, r) for (ll i = l; i < ll(r); i++)
#define rrep(i, l, r) for (ll i = ll(r-1); i >= l; i--)

#pragma region // doubling
template <typename T>
T fpow(T base, T id, ll exp, std::function<T(T, T)> bop);
template <typename T>
T fpow(T base, ll exp);

template <typename T>
T fpow(T base, ll exp) {
    T ans = 1;
    while (exp > 0) {
        if (exp & 1ll) {
            ans = ans * base;
        }
        base = base * base;
        exp >>= 1;
    }
    return ans;
}

template <typename T>
T fpow(T base, T id, ll exp, std::function<T(T, T)> bop) {
    T ans = id;
    while (exp > 0) {
        if (exp & 1ll) {
            ans = bop(ans, base);
        }
        base = bop(base, base);
        exp >>= 1;
    }
    return ans;
}
#pragma endregion


#pragma region
#pragma endregion



#include <atcoder/all>
using namespace atcoder;
using mint = modint998244353;

int main() {
    ll n, a, b, c;
    cin >> n >> a >> b >> c;
    vll ans1(n, LLONG_MAX), ans2(n, LLONG_MAX);
    priority_queue<tuple<ll, ll, bool>, vector<tuple<ll, ll, bool>>, greater<tuple<ll, ll, bool>>> pq;
    pq.emplace(a + b, 1, true);
    pq.emplace(a + b, 1, false);
    set<tuple<ll, ll, bool>> s;
    while (pq.size()) {
        auto [cost, value, addoneable] = pq.top();
       //  cerr << cost << " " << value << " " <<  addoneable << endl;
        pq.pop();
        if (addoneable) {
            if (ans1[value] <= cost) continue;
            ans1[value] = cost;
        }
        else {
            if (ans2[value] <= cost) continue;
            ans2[value] = cost;
        }
        pq.emplace(cost + c, (value * 2) % n, false);
        pq.emplace(cost + a + b + c, (value * 2 + 1) % n, true);
        if (addoneable) pq.emplace(cost + a, (value + 1) % n, true);
    }
    rep(i, 0, n) {
        cout << min(ans1[i], ans2[i]) << "\n";
    }
    cout << flush;
}