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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.StringJoiner;
import java.util.StringTokenizer;
import java.util.function.LongUnaryOperator;

final class Solver {
	static final Solver INSTANCE = new Solver();

	void solve(FastScanner sc) {
		int N = sc.nextInt();
		LongUnaryOperator[] ops = new LongUnaryOperator[N - 1];
		for (int i = 0; i < N - 1; i++) {
			int C = sc.nextInt();
			int S = sc.nextInt();
			int F = sc.nextInt();
			ops[i] = t -> {
				if (S >= t) return C + S;
				else {
					long wait = (t - S) % F;
					if (wait == 0) return t + C;
					return t + (F - wait) + C;
				}
			};
		}
		StringJoiner joiner = new StringJoiner("\n");
		for (int i = 0; i < N; i++) {
			long t = 0;
			for (int j = i; j < N - 1; j++) {
				t = ops[j].applyAsLong(t);
			}
			joiner.add(String.valueOf(t));
		}
		System.out.println(joiner.toString());
	}
}

final class FastScanner extends BufferedReader {
	private StringTokenizer tokenizer;

	FastScanner(InputStream in) {
		super(new InputStreamReader(in));
	}

	String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(readLine());
			} catch (IOException e) {
				throw new UncheckedIOException(e);
			}
		}
		return tokenizer.nextToken();
	}

	int nextInt() {
		return Integer.parseInt(next());
	}

	//10^10~
	long nextLong() {
		return Long.parseLong(next());
	}
}

class Main {
	public static void main(String... args) {
		FastScanner in = new FastScanner(System.in);

		Solver.INSTANCE.solve(in);

		try {
			in.close();
		} catch (IOException e) {
			throw new UncheckedIOException(e);
		}
	}
}