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 <string.h>
#include <stdlib.h>
#include <stdint.h> // int64_t

#define max(a,b) ((a) > (b) ? (a) : (b))
#define min(a,b) ((a) > (b) ? (b) : (a))

int get_int(void) {
  int num;
  scanf("%d", &num);
  return num;
}

int get_int2(int *a1, int *a2) {
  scanf("%d %d", a1, a2);
  return 0;
}

int get_int3(int *a1, int *a2, int *a3) {
  scanf("%d %d %d", a1, a2, a3);
  return 0;
}

#define NUM_MAX 500

struct train {
    int duration;
    int start;
    int interval;
};

int main(void) {
    int num = get_int();

    static struct train trains[NUM_MAX];
    int i;
    for(i = 0; i < num-1; i++) {
        int c, s, f;
        get_int3(&c, &s, &f);
        trains[i] = (struct train){c, s, f};
    }

    static int ans[NUM_MAX];
    int start;
    for(start = num-2; start >= 0; start--) {
        int elapse = 0;
        for(i = start; i < num-1; i++) {
            struct train t = trains[i];
            if(elapse <= t.start) {
                elapse = t.start;
            } else {
                if(elapse%t.interval) {
                    elapse = (elapse/t.interval+1)*t.interval;
                }
            }
            elapse += t.duration;
#ifdef DEBUG
            printf("start: %d -> %d\n", start, elapse);
#endif
        }
        ans[start] = elapse;
    }

    for(i = 0; i < num; i++) {
        printf("%d\n", ans[i]);
    }
    return 0;
}