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)

use std::io;


fn read_line() -> String {
    let mut in_str = String::new();
    io::stdin().read_line(&mut in_str).unwrap();
    return in_str
}

fn main(){


    const MAX_N : usize = 501;

    let ln1 = read_line();
    let n : usize = (ln1.trim().parse::<i32>().unwrap() - 1) as usize;

    let mut cs : [i32 ; MAX_N] = [0 ; MAX_N];
    let mut ss : [i32 ; MAX_N] = [0 ; MAX_N];
    let mut fs : [i32 ; MAX_N] = [0 ; MAX_N];

    for i in 0..n {
        let ln2 = read_line();
        let v : Vec<i32> = ln2.split_whitespace().map(|x| x.parse::<i32>().unwrap()).collect::<Vec<i32>>();
        cs[i] = v[0];
        ss[i] = v[1];
        fs[i] = v[2];
    }    


    for j in 0..n {
        let mut elappsed : i32 = 0;
        for i in j..n {
            if ss[i] >= elappsed {
                elappsed = ss[i] + cs[i];
            }
            else{
                elappsed = (elappsed + fs[i] - 1) / fs[i] * fs[i] + cs[i];
            }
        }
        println!("{}", elappsed);
    }
    println!("{}", 0);
}