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)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) for (int i = (0); i < (n); ++i)
#define ALL(obj) (obj).begin(), (obj).end()
#define DEBUG(val) cout << #val << " : " << val << endl
#define MOD 1000000007

int n;
int min_dist[505];
int ans[505];
vector<tuple<int, int, int>> data;

int pre_station_min_time(int station, int start_station) {
  if (station == start_station) {
    tuple<int, int, int> station_data = data[start_station - 1];
    return (get<0>(station_data) + get<1>(station_data));
  }
  tuple<int, int, int> station_data = data[station - 1];
  int prev_time = pre_station_min_time(station - 1, start_station);
  if (prev_time <= get<1>(station_data)) {
    return max(prev_time + get<0>(station_data),
               get<0>(station_data) + get<1>(station_data));
  } else {
    int base_time = prev_time - get<1>(station_data);
    int over_time =
        round(base_time / (float)get<2>(station_data)) * get<2>(station_data);
    int next_time = over_time + get<1>(station_data) + get<0>(station_data);
    return next_time;
  }
}

int main() {
  cin >> n;
  REP(i, n - 1) {
    int c, s, f;
    cin >> c >> s >> f;
    tuple<int, int, int> t;
    t = make_tuple(c, s, f);
    data.push_back(t);
  }
  FOR(i, 1, n + 1) {
    if (i == n) {
      cout << 0 << endl;
    } else {
      cout << pre_station_min_time(n - 1, i) << endl;
    }
  }
}