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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

#!/mnt/c/Users/moiki/bash/env/bin/python
# N,M = map(int, input().split())
N = int(input())
trains = [ [int(j) for j in input().split()] for _ in range(N-1)] + [(0,0,0)]
dp = [float("inf") for _ in range(N-1)] + [0] # N-1

t = trains[N-2]
dp[N-2] = t[0]+t[1]
limited_by = [0 for _ in range(N)]
limited_by[N-2] = t[1]

for i in range(N-3, -1, -1):
    c,s,f = trains[i]
    nc,ns,nf = trains[i+1]
    arrive_at = c+s # start with A : B
    # limited_by[i] = max(limited_by[i+1]-c, s)
    
    # if arrive_at <

    if arrive_at < limited_by[i+1]-c:
        dp[i] = dp[i+1]
        limited_by[i] = limited_by[i+1] -c 

    else:
        dp[i] = arrive_at+ (f - arrive_at%f)%f + dp[i+1] - limited_by[i+1]
        limited_by[i] = s

for i in range(N): print(dp[i])
# print(limited_by)



# print(dp)
# 
# 
# for i in range(N-3, -1, -1):
#     c,s,f = trains[i]
#     nc,ns,nf = trains[i+1]
#     arrive_at = c+s
#     # print("i: ", i, "c,s,f: ", c,s,f, end=" ")
#     # print("arrive_at: ", arrive_at, end=" ")
#     # print(" arrive_at<ns: ", arrive_at < ns)
# 
#     if arrive_at < ns:
#         dp[i] = dp[i+1]
#         limited_by[i] = ns - c
#     elif arrive_at < limited_by[i+1]:
#         dp[i] = dp[i+1]
#         limited_by[i] = limited_by[i+1] - c
#     else:
#         # print("arrive_at: ", arrive_at, "f: ", f, "dp[i+1] : ", dp[i+1], "ns: ", ns)
#         dp[i] = arrive_at+ (f - arrive_at%f)%f + dp[i+1] - ns
#