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)

#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<string>
#include<iostream>
#include<iostream>
#include<cctype>
#include<cstdio>
#include<vector>
#include<stack>
#include<queue>
#include <algorithm>
#include<math.h>
#include<set>
#include<map>
#include <sstream>
#include<iomanip>
#include <ctype.h>
using namespace std;

//#include<bits/stdc++.h>


int main() {
	int n;
	vector<int>a, b, c;
	cin >> n;
	vector<long long int>w;
	long long int t = 0;
	//w.push_back(t);
	int x, y, z;
	//a.push_back(0);
	//b.push_back(0);
	//c.push_back(0);
	for (int i = 0; i < n-1; i++) {
		cin >> x >> y >> z;
	//	if (t){
			if (t <= y)t = y;
			else if ((t - y) % z) {
				long long int r = (t - y) % z+1;
				t = y + z*r;
			}
			//cout << t;
	//	}
		t += x;

		//w.push_back(t);
		a.push_back(x);
		b.push_back(y);
		c.push_back(z);
		
	}
//	w.push_back(t);
	
	for (int i = 0; i <n-1; i++)
	{
		t = 0;
		
		for (int j = i; j <n-1; j++) {
			x = a[j]; y = b[j]; z = c[j];
			if (t < y)t = y;
			else if (t % z) {
				t += z - t%z;
			}
			t += x;
		}
		w.push_back(t);
		
	}
	w.push_back(0);
	sort(w.begin(), w.end());
	reverse(w.begin(), w.end());
	for (int i = 0; i <w.size(); i++)
		cout << w[i] << endl;

	return 0;
}