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 Ruby (2.3.3) (judge verdict: Wrong Answer)

n = gets.to_i
trains = []
(n-1).times do
  trains << gets.chomp.split(" ").map(&:to_i)
end

# trains[i][0] : C
# trains[i][1] : S
# trains[i][2] : F

# 駅0に居る場合にどうするか、
# 駅1に居る場合
# …
# 駅nに居る場合

time = Array.new(n,0)

(n-1).times do |i|
  # 最初に来る時間と次の駅までの時間
  time[i] += trains[i][1] + trains[i][0]

#  puts "初期値：#{time}"
  (i+1).upto(n-2) do |j|
    # 待ち時間
    if time[i] < trains[j][1]
      time[i] += trains[j][1] - time[i]
    elsif time[i] > trains[j][1]
      rest = time[i] - trains[j][1]
      b = rest / trains[j][2]
      if rest % trains[j][2] != 0
        b += 1
      end
      nxt = trains[j][1] + (trains[j][2] * (b))
      time[i] += time[i] - nxt
    end
#    puts "rest : #{time}"
    # 次の駅までの時間
    time[i] += trains[j][0]
#    puts "駅#{j}追加： #{time}"
  end

end

puts time