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 (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {
    val inputN = readLine().orEmpty().toInt()
    val diagram = mutableListOf<Triple<Int,Int,Int>>()
    val ansList = mutableListOf<Int>()

    for(i in 0 until inputN - 1){
        val dia = readLine()?.split(" ").orEmpty()

        val c = dia[0].toInt()
        val s = dia[1].toInt()
        val f = dia[2].toInt()
        diagram.add(Triple(c,s,f))
    }
    var j = inputN -1
    var k = 0
    while(j > - 1){
        if(j == inputN - 1){
            ansList.add(0)
        }else if(j == inputN - 2){
            ansList.add(diagram[j].second + diagram[j].first)
        }else{
            val nextWaitTime = diagram[j + 1].second
            val nextInterval = diagram[j + 1].third
            var time = diagram[j].second + diagram[j].first

            if(time <= nextWaitTime){
                time = ansList[k - 1]
            }else{
                var delay = (time - nextWaitTime) / nextInterval
                if((time - nextWaitTime) % nextInterval == 0){
                    time = ansList[k - 1] + (nextInterval * delay)
                }else{
                   delay += 1
                    time = ansList[k - 1] + (nextInterval * delay)
                }
            }
            ansList.add(time)
        }
        j -= 1
        k += 1
    }

    val showList = ansList.reversed()
    showList.forEach{
        println(it)
    }

}