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 Kotlin (1.3.71) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>){

    //var (N,M) = readLine()!!.split(" ").map{it.toInt()}
    var N = readLine()!!.toInt()
    var C = LongArray(N)
    var S = LongArray(N)
    var F = LongArray(N)
    var ans = LongArray(N){0}

    for(i in 0 until N-1){
        var (ci,si,fi) = readLine()!!.split(" ").map{it.toLong()}
        C[i] = ci
        S[i] = si
        F[i] = fi
    }

    ans[N-2] = S[N-2] + C[N-2]

    for(i in 0 until N-1){
        var time = 0L
        for(j in i until N-1){
            when(time <= S[j]){
                true -> time += (S[j] - time) + C[j]
                false -> time += time % F[j] + C[j]
            }
        }
        println(time)
    }
    println("0")

//    for(i in N-3 downTo 0){
//        if(S[i]+C[i] <= S[i+1]){
//            ans[i] = ans[i+1]
//        } else {
//            ans[i] = (ans[i+1] - S[i+1]) + S[i] + C[i] + ((S[i] + C[i]) % F[i+1])
//        }
//    }
//    for(i in 0 until N){
//        println(ans[i])
//    }

}