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)

fn main() {
    let n = readln();
    let v: Vec<(i32, i32, i32)> = readlns(n-1);
    for i in 0..n {
        let mut cur = 0;
        for &(c, s, f) in &v[i..n-1] {
            cur = max(s, (cur + f-1) / f * f) + c;
        }
        println!("{}", cur);
    }
}

// --- template ---
#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::HashSet;

pub trait FromLn {
    fn fromln(s: &str) -> Self;
}
pub fn readln<T: FromLn>() -> T {
    let mut buf = String::new();
    let _ = ::std::io::stdin().read_line(&mut buf).unwrap();
    T::fromln(buf.trim())
}
pub fn readlns<T: FromLn>(n: usize) -> Vec<T> {
    let mut vs = vec![];
    for _ in 0..n { vs.push(readln()); }
    vs
}
macro_rules! fromln_primitives {
    ($($t:ty),*) => { $(
        impl FromLn for $t {
            fn fromln(s: &str) -> $t {
                s.parse().unwrap()
            }
        }
    )* }
}
fromln_primitives!(String, bool, f32, f64, isize, i8, i16, i32, i64, usize, u8, u16, u32, u64);
impl<T> FromLn for Vec<T> where T: FromLn {
    fn fromln(s: &str) -> Vec<T> {
        s.split_whitespace().map(T::fromln).collect()
    }
}
impl FromLn for Vec<char> {
    fn fromln(s: &str) -> Vec<char> {
        s.chars().collect()
    }
}
macro_rules! fromln_tuple {
    ($($t:ident),*) => {
        impl<$($t),*> FromLn for ($($t),*) where $($t: FromLn),* {
            fn fromln(s: &str) -> ($($t),*) {
                let mut it = s.split_whitespace();
                let t = ($($t::fromln(it.next().unwrap())),*);
                assert_eq!(it.next(), None);
                t
            }
        }
    }
}
fromln_tuple!(A, B);
fromln_tuple!(A, B, C);
fromln_tuple!(A, B, C, D);
fromln_tuple!(A, B, C, D, E);
fromln_tuple!(A, B, C, D, E, F);