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::*;
#[allow(unused_imports)]
use std::str::*;
#[allow(unused_imports)]
use std::collections::*;

#[allow(dead_code)]
fn read<T: FromStr>(s: &mut StdinLock) -> T {
    let s = s.by_ref().bytes().map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect::<String>();
    s.parse::<T>().ok().unwrap()
}

#[allow(unused_macros)]
macro_rules! p {
    ($x:expr) 
    => (println!("{}", $x));
    ($x:expr, $y:expr)
    => (println!("{} {}", $x, $y));
    ($x:expr, $y:expr, $z:expr) 
    => (println!("{} {} {}", $x, $y, $z));
}

#[allow(unused_macros)]
macro_rules! pl {
    ($x:expr) 
    => (println!("{}", $x));
    ($x:expr, $y:expr)
    => (println!("{}\n{}", $x, $y));
    ($x:expr, $y:expr, $z:expr) 
    => (println!("{}\n{}\n{}", $x, $y, $z));
}

fn main() {
    let s = stdin();
    let mut s = s.lock();
    let S = &mut s;

    for _ in 0..1 {
        let n: usize = read(S);

        let mut c = vec![]; // かかる時間
        let mut s = vec![]; // 開始時間
        let mut f = vec![]; // 周期
        for i in 0..n-1 {
            let x: i32 = read(S);
            c.push(x);
            let x: i32 = read(S);
            s.push(x);
            let x: i32 = read(S);
            f.push(x);
        }

        for i in 0..n {
            let mut cur_t = 0;
            let mut j = i;
            while j != n - 1 {
                if cur_t < s[j] {
                    cur_t = s[j];
                }
                if (cur_t - s[j]) % f[j] != 0 {
                    cur_t += f[j] - (cur_t - s[j]) % f[j];
                }
                cur_t += c[j];
                j += 1;
            }
            p!(cur_t);
        }
    }
}