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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>

int max(int a, int b){
	return a >= b ? a : b;
}

int main(){
	int N, i, j;
	scanf("%d", &N);
	int *C = (int *)malloc(sizeof(int) * (N - 1));
	int *S = (int *)malloc(sizeof(int) * (N - 1));
	int *F = (int *)malloc(sizeof(int) * (N - 1));
	int *ans = (int *)malloc(sizeof(int) * N);
	for(i = 0; i < N - 1; i++){
		scanf("%d%d%d", &C[i], &S[i], &F[i]);
		ans[i] = 0;
	}
	ans[N - 1] = 0;
	for(i = 0; i < N - 1; i++){
		for(j = i; j < N - 1; j++){
			ans[i] = max(S[j], ((ans[i] + F[j] - 1) / F[j]) * F[j]);
//			printf("%d\n", ans[i]);
			ans[i] += C[j];
//			printf("(i, j) = (%d, %d):%d\n", i, j, ans[i]);
		}
	}
	for(i = 0; i < N; i++){
		printf("%d\n", ans[i]);
	}
	return 0;
}