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_imports)]
use std::io::{self, BufRead};
#[allow(unused_imports)]
use std::cmp::{min, max};

#[allow(dead_code)]
fn readln_to_vec<T>(buf: &mut BufRead) -> Vec<T> 
where T: std::str::FromStr,
      <T as std::str::FromStr>::Err: std::fmt::Debug
{
    let mut s = String::new();
    buf.read_line(&mut s).unwrap();
    s.split(" ")
        .map(|x| x.trim())
        .filter(|x| !x.is_empty())
        .filter_map(|x| x.parse().ok())
        .collect()
}

#[allow(dead_code)]
fn readln_to<T>(buf: &mut BufRead) -> T
where T: std::str::FromStr + std::clone::Clone,
      <T as std::str::FromStr>::Err: std::fmt::Debug
{
    readln_to_vec::<T>(buf)[0].clone()
}

#[allow(unused_macros)]
macro_rules! readln_to_tuple {
    (
        $bufRead: expr, ( $( $t: ty ),* )
    ) => (
        {
            let buf: &mut BufRead = $bufRead;
            let ln = readln_to_vec::<String>(buf);
            let mut i = ln.into_iter();
            ( $( i.next().unwrap().parse::<$t>().unwrap() ),* )
        }
        )
}

fn solve(t: i64, k: usize, n: usize, c: &Vec<i64>, s: &Vec<i64>, f: &Vec<i64>) -> i64 {
    if k == n-1 {
        t
    } else {
        if t < s[k] {
            solve(s[k] + c[k], k+1, n, c, s, f)
        } else {
            solve(((t + f[k]-1) / f[k]) * f[k] + c[k], k+1, n, c, s, f)
        }
    }
}


fn main() {
    let stdin = io::stdin();
    let mut stdin = stdin.lock();
    let stdin = &mut stdin;

    let n: usize = readln_to(stdin);

    let mut c = Vec::new();
    let mut s = Vec::new();
    let mut f = Vec::new();

    for _ in 0..n-1 {
        let (ci, si, fi) = readln_to_tuple!(stdin, (i64, i64, i64));
        c.push(ci);
        s.push(si);
        f.push(fi);
    }

    for i in 0..n {
        println!("{}", solve(0, i, n, &c, &s, &f));
    }
}