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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define vvi vector<vector<int>>
#define vec vector
#define pq priority_queue
#define all(v) (v).begin(), (v).end()
#define uniqueV(x) sort(x.begin(), x.end()); x.erase(unique(x.begin(), x.end()), x.end());
#define rep(i, n) for (int (i) = (0); (i) < (n); ++(i))
#define repp(i, m, n) for (int (i) = (m); (i) < (n); ++(i))
#define debug(x) cerr << #x << ": " << x << endl;
#define debug2(x, y) cerr<<"("<<#x<<", "<<#y<<") = "<<"("<<x<<", "<<y<<")"<<endl;
#define debug3(x, y, z) cerr<<"("<<#x<<", "<<#y<<", "<<#z<<") = "<<"("<<x<<", "<<y<<", "<<z<<")"<<endl;
#define debugB(value, size) cerr<<#value<<": "<<bitset<size>(value) << endl;
#define line() cerr << "---------------" << endl;

const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, -1, 1};
const double PI = 3.14159265358979323846;

template<typename T>
void printA(vector<T> &printArray, char between = ' ') {
   int paSize = printArray.size();
   for (int i = 0; i < paSize; i++) { cerr << printArray[i] << between;}
   if (between != '\n') {cerr << endl;}
}

// ------------------------------------------------------------------------------------------

int N;
int C[555], S[555], F[555];

int main() {
   cin >> N;
   rep (i, N-1) {
      cin >> C[i] >> S[i] >> F[i];
   }

   rep (i, N) { // スタート
      int t = 0; // 今の時間
      repp (j, i, N-1) {
         if (t < S[j]) {
            t = S[j];
         } else if (t % F[j] == 0) {

         } else {
            t = t + F[j] - t % F[j];
         }
         t += C[j];
      }
      cout << t << endl;
   }

   return 0;
}