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;

int main(int argc, char *argv[]) {
    int N;
    cin >> N;

    if (N == 1) {
    } else {
        vector<int> carr(N - 1, 0);
        vector<int> sarr(N - 1, 0);
        vector<int> farr(N - 1, 0);
        for (int i = 0; i < N - 1; i++) {
            int c, s, f;
            cin >> c >> s >> f;
            carr[i] = c;
            sarr[i] = s;
            farr[i] = f;
        }

        for (int j = 1; j < N; j++) {
            int idx = j - 1;
            int ans = 0;
            int start = carr[idx] + sarr[idx];
            for (int k = idx; k < N - 1; k++) {
                if (k == idx) {
                    ans += start;
                } else {
                    //cout << "sarr[k] " << sarr[k] << endl;
                    //cout << "farr[k] " << farr[k] << endl;
                    //cout << "carr[k] " << carr[k] << endl;
                    //cout << ((ans / sarr[k]) * sarr[k]) + farr[k] + carr[k] << endl;
                    int start = sarr[k];
                    if (ans >= sarr[k]) {
                        start = ((ans / sarr[k]) * sarr[k]);
                        if (ans % sarr[k] != 0) {
                            start += farr[k];
                        }
                    }
                    //cout << "start:" << start << endl;;
                    ans = start + carr[k];
                }
                //cout << "ans:" << ans << endl;
            }
            cout << ans << endl;
        }
    }
    cout << 0 << endl;
   
    return 0;
}