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 <iostream>
#include <vector>
#include <fstream>
#include <algorithm> // sort
#include <map> // pair
using namespace std;
#define REP(i, n) for (long long i = 0; i < (n); i++) 
typedef long long ll;
static const ll MOD = 1000000007;
static const ll INF = 1000000000000000000LL;
using namespace std;


//後ろから求める
//最短時間だから
int main() {
    int N;
    cin >> N;

    int c[N-1],s[N-1],f[N-1];

    REP(i,N-1)
    {
      cin >> c[i] >> s[i] >> f[i];
    }

    //res[N-1]からもとめる
    ll res[N+1] = {};

    for(int i = N-2; i >= 0; --i)
    {
      int cnt = 0;

      cnt += s[i] + c[i];
      for(int j = i; j < N-2; ++j)
      {
        int tmp;
        //s%f == 0なので
        tmp = cnt % f[j+1];

        //乗れる電車を探す
        cnt += tmp;
        if(cnt < s[j+1])
        {
          cnt = s[j+1];
        }
        cnt += c[j+1];
      }

      res[i] = cnt;
    }
    REP(i,N)
    {
      cout << res[i] << endl;
    }




    return 0;
}