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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define PI 3.14159265359
#define NIL (-1)
#define LL long long
using namespace std;
const int64_t MOD = 1e9 + 7;

int main() {
	int N;
	cin >> N;

	struct info {
		int C;
		int S;
		int F;
	};

	vector<struct info> v;
	for (int i = 0; i < N - 1; i++) {
		struct info Info;
		int c, s, f;
		cin >> c >> s >> f;
		//cout << c << s << f << endl;

		Info = { c,s,f };
		v.push_back(Info);
	}

	//for (int i = 0; i < N - 1; i++) {
	//	cout << v.at(i).C << ' ';
	//	cout << v.at(i).S << ' ';
	//	cout << v.at(i).F << endl;;
	//}

	vector<int> A(N);
	for (int i = 0; i < N - 1; i++) {
		int sum = 0;

		for (int j = i; j < N - 1; j++) {
			if (sum < v.at(j).S) sum = v.at(j).S;
			else if (!(sum % v.at(j).F == 0)) sum = sum + v.at(j).F - (sum % v.at(j).F);
			sum += v.at(j).C;
		}

		A.at(i) = sum;
	}

	for (int i = 0; i < N; i++) {
		cout << A.at(i) << endl;
	}
}