Problem:
Score : 300 points

Problem Statement

A railroad running from west to east in Atcoder Kingdom is now complete.

There are N stations on the railroad, numbered 1 through N from west to east.

Tomorrow, the opening ceremony of the railroad will take place.

On this railroad, for each integer i such that 1≤i≤N-1, there will be trains that run from Station i to Station i+1 in C_i seconds. No other trains will be operated.

The first train from Station i to Station i+1 will depart Station i S_i seconds after the ceremony begins. Thereafter, there will be a train that departs Station i every F_i seconds.

Here, it is guaranteed that F_i divides S_i.

That is, for each Time t satisfying S_i≤t and t％F_i=0, there will be a train that departs Station i t seconds after the ceremony begins and arrives at Station i+1 t+C_i seconds after the ceremony begins, where A％B denotes A modulo B, and there will be no other trains.

For each i, find the earliest possible time we can reach Station N if we are at Station i when the ceremony begins, ignoring the time needed to change trains.

Constraints

1≤N≤500

1≤C_i≤100

1≤S_i≤10^5

1≤F_i≤10

S_i％F_i=0

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
C_1 S_1 F_1
:
C_{N-1} S_{N-1} F_{N-1}

Output

Print N lines. Assuming that we are at Station i (1≤i≤N) when the ceremony begins, if the earliest possible time we can reach Station N is x seconds after the ceremony begins, the i-th line should contain x.

Sample Input 1

3
6 5 1
1 10 1

Sample Output 1

12
11
0

We will travel from Station 1 as follows:

5 seconds after the beginning: take the train to Station 2.

11 seconds: arrive at Station 2.

11 seconds: take the train to Station 3.

12 seconds: arrive at Station 3.

We will travel from Station 2 as follows:

10 seconds: take the train to Station 3.

11 seconds: arrive at Station 3.

Note that we should print 0 for Station 3.

Sample Input 2

4
12 24 6
52 16 4
99 2 2

Sample Output 2

187
167
101
0

Sample Input 3

4
12 13 1
44 17 17
66 4096 64

Sample Output 3

4162
4162
4162
0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS
#include "bits/stdc++.h"
#include <random>

using namespace std;

//呪文
#define DUMPOUT cerr
#define dump(...) DUMPOUT<<"  ";DUMPOUT<<#__VA_ARGS__<<" :["<<__LINE__<<":"<<__FUNCTION__<<"]"<<endl;DUMPOUT<<"    ";dump_func(__VA_ARGS__)

typedef unsigned uint; typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef pair<double, double> pdd; typedef pair<string, string> pss;
template <typename _KTy, typename _Ty> ostream& operator << (ostream& ostr, const pair<_KTy, _Ty>& m) { ostr << "{" << m.first << ", " << m.second << "}"; return ostr; }
template <typename _KTy, typename _Ty> ostream& operator << (ostream& ostr, const map<_KTy, _Ty>& m) { if (m.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << *m.begin(); for (auto itr = ++m.begin(); itr != m.end(); itr++) { ostr << ", " << *itr; } ostr << "}"; return ostr; }
template <typename _Ty> ostream& operator << (ostream& ostr, const vector<_Ty>& v) { if (v.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << v.front(); for (auto itr = ++v.begin(); itr != v.end(); itr++) { ostr << ", " << *itr; }	ostr << "}"; return ostr; }
template <typename _Ty> ostream& operator << (ostream& ostr, const set<_Ty>& s) { if (s.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << *(s.begin()); for (auto itr = ++s.begin(); itr != s.end(); itr++) { ostr << ", " << *itr; }	ostr << "}"; return ostr; }
template <typename _Ty> ostream& operator << (ostream& ostr, const stack<_Ty>& s) { if (s.empty()) { ostr << "{ }"; return ostr; } stack<_Ty> t(s); ostr << "{" << t.top(); t.pop(); while (!t.empty()) { ostr << ", " << t.top(); t.pop(); } ostr << "}";	return ostr; }
template <typename _Ty> ostream& operator << (ostream& ostr, const list<_Ty>& l) { if (l.empty()) { ostr << "{ }"; return ostr; } ostr << "{" << l.front(); for (auto itr = ++l.begin(); itr != l.end(); ++itr) { ostr << ", " << *itr; } ostr << "}"; return ostr; }
template <typename _KTy, typename _Ty> istream& operator >> (istream& istr, pair<_KTy, _Ty>& m) { istr >> m.first >> m.second; return istr; }
template <typename _Ty> istream& operator >> (istream& istr, vector<_Ty>& v) { for (size_t i = 0; i < v.size(); i++) istr >> v[i]; return istr; }
namespace aux { // print tuple
	template<typename Ty, unsigned N, unsigned L> struct tp { static void print(ostream& os, const Ty& v) { os << get<N>(v) << ", "; tp<Ty, N + 1, L>::print(os, v); } };
	template<typename Ty, unsigned N> struct tp<Ty, N, N> { static void print(ostream& os, const Ty& value) { os << get<N>(value); } };
}
template<typename... Tys> ostream& operator<<(ostream& os, const tuple<Tys...>& t) { os << "{"; aux::tp<tuple<Tys...>, 0, sizeof...(Tys)-1>::print(os, t); os << "}"; return os; }

template<typename A, size_t N, typename T> inline void Fill(A(&array)[N], const T &val) { std::fill((T*)array, (T*)(array + N), val); }

void dump_func() { DUMPOUT << endl; }
template <class Head, class... Tail> void dump_func(Head&& head, Tail&&... tail) { DUMPOUT << head; if (sizeof...(Tail) == 0) { DUMPOUT << " "; } else { DUMPOUT << ", "; } dump_func(std::move(tail)...); }

#define PI 3.14159265358979323846
#define EPS 1e-11
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define all(x) (x).begin(), (x).end()
#define SZ(x) ((int)(x).size())
#define fake false



int main() {

	random_device seed_gen;
	mt19937 engine;
	engine.seed(seed_gen());

	cin.tie(0);
	ios::sync_with_stdio(false);

	int N;
	cin >> N;
	vector<int> C(N - 1), S(N - 1), F(N - 1);
	REP(i, N - 1) {
		cin >> C[i] >> S[i] >> F[i];
	}

	// Si + n * Fi

	for (int i = 0; i < N - 1; i++) {
		int t = S[i];
		for (int j = i; j < N - 1; j++) {	
			t += C[j];
			// 駅 j + 1 に到着
			if (j + 1 == N - 1) break;
			t = max(S[j + 1], (t % F[j] == 0 ? t : t + (F[j] - t % F[j])));
		}
		cout << t << endl;
	}
	cout << 0 << endl;


	return 0;
}