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<iostream>
#include<string>
#include<vector>
#include<stdio.h>
#include<algorithm>
#include<math.h>
#include<numeric>
#include<iomanip>
#include<deque>
#include<tuple>
#include<queue>
#include<map>
#include<cstdint>
#include<unordered_map>
#include<boost/multiprecision/cpp_int.hpp>


#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define vi vector<int>
#define all(x) (x).begin(),(x).end()
#define Endl endl
#define F first
#define S second

namespace mp = boost::multiprecision;

using cpp_int = mp::cpp_int;
using ll = long long;
using namespace std;


int main() {
	int n;
	cin >> n;
	vector<int>C(n), S(n), F(n);
	rep(i, n-1) {
		cin >> C[i] >> S[i] >> F[i];
	}
	rep(i, n) {
		int time = S[i] + C[i];
		FOR(j, i+1, n - 1) {
			int buffer = 0;
			if (time > S[j]) {
				time -= S[j];
				if (time % F[j] == 0) {
					buffer = time / F[j];
				}
				else {
					buffer = time / F[j] + 1;
				}
				buffer *= F[j];
			}
			time = buffer + S[j] + C[j];
		}
		cout << time << endl;
	}
}