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: Accepted)

data class Data(val c: Int, val s: Int, val f: Int)

fun main(args: Array<String>) {
    val n = readLine()?.toInt() ?: return
    val datas: List<Data> = (0 until n - 1).map {
        val (c, s, f) = readLine()?.split(" ")?.map { it.toInt() }!!
        Data(c, s, f)
    }

    // n駅については最後に0を出力すればよいので考えない
    for (i in datas.indices) {
        var total = datas[i].s + datas[i].c // 出発時間 + 運行時間

        // n-1駅についてはn駅より先の駅について考える必要はない
        if (i == datas.size - 1) {
            println(total)
            break
        }

        // 1 ~ n-2駅については、n-1までループして時間計算する
        for (j in (i + 1) until datas.size) {
            // j駅に到着してもj駅から電車が出ていない場合
            if (datas[j].s >= total) {
                total = datas[j].s + datas[j].c
                continue
            }

            // j駅に到着してからの待ち時間を計算する
            val diff = total - datas[j].s
            val remainder = diff % datas[j].f
            val waitTime = if (remainder == 0) 0 else datas[j].f - remainder
            total += (waitTime + datas[j].c)
        }

        println(total)
    }
    println(0)
}