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 secs = 0, station = 0, transit=0, n;
vector<int> c, s, f, wait;
 
void step(int t) {
	// t is either wait[station] or transit
	secs+=t;
	cout << t << ' ';
	if (t == transit) {
		wait[station]-=secs;
		transit=0;
	}
	else wait[station]=0;

	if (wait[station]<=0)
		wait[station] = abs(wait[station])%f[station];
	if (wait[station]==0) {
		transit=c[station];
		station++;
	}
}
 
void reset() {
	secs=0;
	station=0;
	transit=0;
	for (int i=0; i<n; i++)
		wait[i]=s[i];
}
 
int main() {
	cin>>n;
	c.resize(n);
	s.resize(n);
	f.resize(n);
	wait.resize(n);
 
	for (int i=0; i<n-1; i++)
		cin>>c[i]>>s[i]>>f[i];
 
	//simulate v9
	for (int i=0; i<n; i++) {
		reset();
		station=i;
		while (station < n-1) {
			if (transit > 0) step(transit);
			else step(wait[station]);
		}
		cout << secs+transit << endl;
	}
 
	return 0;
}