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)

#[allow(unused_macros)]
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 ( 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 ) * } } ; }
#[allow(unused_macros)]
macro_rules ! input_inner { ( $ next : expr ) => { } ; ( $ next : expr , ) => { } ; ( $ next : expr , mut $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let mut $ var = read_value ! ( $ next , $ t ) ; input_inner ! { $ next $ ( $ r ) * } } ; ( $ next : expr , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ next , $ t ) ; input_inner ! { $ next $ ( $ r ) * } } ; }
#[allow(unused_macros)]
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: [(u64, u64, u64); n-1]
    }

    for i in 0..n {
        let mut k = i;
        let mut t = 0;
        while k != n-1 {
            let (c, s, f) = csf[k];
            if t < s {
                t = s + c;
                k += 1;
            } else {
                t = s + ceil_div(t - s, f) * f + c;
                k += 1;
            }
        }
        println!("{}", t);
    }
}

#[allow(dead_code)]
fn ceil_div(x: u64, y: u64) -> u64 {
    (x + y - 1) / y
}