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 Go (1.6) (judge verdict: Accepted)

package main

import "fmt"

func main() {
	var N int
	fmt.Scan(&N)
	C := make([]int, N-1)
	S := make([]int, N-1)
	F := make([]int, N-1)
	for i := 0; i < N-1; i++ {
		fmt.Scan(&C[i])
		fmt.Scan(&S[i])
		fmt.Scan(&F[i])
	}
	for i := 0; i < N-1; i++ { // 駅0~N-2まで(0-indexed)
		// 駅iを何時に出発できるか？ S[i]+xF[i] >= now を満たす最小のxを求める。
		// nowを更新する
		now := S[i] + C[i]
		// fmt.Printf("i %d, now %d\n", i, now)
		for j := i + 1; j < N-1; j++ { // 駅iからN-2駅まで
			// 時刻nowで、j駅にいる
			// 次に出発できるのは何時か？
			x := 0
			for {
				if S[j]+x*F[j] >= now {
					break
				}
				x++
			}
			now += (S[j] + x*F[j] - now) + C[j]
			// fmt.Printf("i %d, j %d, x %d, now %d\n", i, j, x, now)
		}
		fmt.Println(now)
	}
	fmt.Println(0)
}