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

c, s, f = [], [], []
(n-1).times do
  ci, si, fi = gets.chomp.split.map(&:to_i)
  c << ci
  s << si
  f << fi
end
c << 0
s << 1
f << 1
# なんとなくシミュレーションする
# 終点だったら終了
# (次の待ち時間)+次への乗車時間
# - 次の発車時間 == 到着時間 -> すぐ乗れる
# - 到着時間 < 次の始発 -> 始発に乗る
# - それ以外（待ち時間） -> 到着時間 + （到着時間-(到着時間/発車間隔+1)*発車間隔）
n.times do |i|
  now = 0
  (i...n-1).each do |j|
    if now < s[j]
      now = s[j]
    elsif (now % f[j]).zero?
      # なにもしない
    else
      # now += now-(now/f[j+1]+1)*f[j+1]
      now = now + f[j+1]-now%f[j+1] # (now/f[j+1]+1)*f[j+1]
    end
    now += c[j]
  end
  puts now
end