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 Rust (1.15.1) (judge verdict: Accepted)

macro_rules !input {(source =$s :expr ,$($r :tt ) *) =>{let mut iter =$s .split_whitespace () ;let mut next =||{iter .next () .unwrap () } ;input_inner !{next ,$($r ) *} } ;($($r :tt ) *) =>{let stdin =std ::io ::stdin () ;let mut bytes =std ::io ::Read ::bytes (std ::io ::BufReader ::new (stdin .lock () ) ) ;let mut next =move ||->String {bytes .by_ref () .map (|r |r .unwrap () as char ) .skip_while (|c |c .is_whitespace () ) .take_while (|c |!c .is_whitespace () ) .collect () } ;input_inner !{next ,$($r ) *} } ;}
macro_rules !input_inner {($next :expr ) =>{} ;($next :expr ,) =>{} ;($next :expr ,$var :ident :$t :tt $($r :tt ) *) =>{let $var =read_value !($next ,$t ) ;input_inner !{$next $($r ) *} } ;}
macro_rules !read_value {($next :expr ,($($t :tt ) ,*) ) =>{($(read_value !($next ,$t ) ) ,*) } ;($next :expr ,[$t :tt ;$len :expr ] ) =>{(0 ..$len ) .map (|_ |read_value !($next ,$t ) ) .collect ::<Vec <_ >>() } ;($next :expr ,chars ) =>{read_value !($next ,String ) .chars () .collect ::<Vec <char >>() } ;($next :expr ,usize1 ) =>{read_value !($next ,usize ) -1 } ;($next :expr ,$t :ty ) =>{$next () .parse ::<$t >() .expect ("Parse error" ) } ;}

fn main() {
   input! {
      n: usize,
      csf: [(usize, usize, usize); n-1],
   }
   for i in 0..n-1 {
      let (c0, s0, _) = csf[i].clone();
      let mut now = s0 + c0;
      for j in i+1..n-1 {
         let (c, s, f) = csf[j].clone();
         if now < s {
            now = s;
         } else if now % f != 0 {
            now += f - now % f;
         }
         now += c;
      }
      println!("{}", now);
   }
   println!("0");
}