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::{stdin, Read, StdinLock};
#[allow(unused_imports)]
use std::cmp::{max, min, Ordering};
#[allow(unused_imports)]
use std::str::FromStr;
#[allow(unused_imports)]
use std::collections::{HashSet, HashMap, BinaryHeap};
#[allow(unused_imports)]
use std::vec::Vec;

#[allow(dead_code)]
const INF: i32 = 1000_000_000;
#[allow(dead_code)]
const INFLL: i64 = 1000_000_000_000_000_000;
#[allow(dead_code)]
const EPS: f64 = 1.0e-10;
#[allow(dead_code)]
const MOD: i32 = 1000_000_007;

struct Scanner<'a> {
    cin: StdinLock<'a>,
}

impl<'a> Scanner<'a> {
    #[allow(dead_code)]
    fn new(cin: StdinLock<'a>) -> Scanner<'a> {
        Scanner {cin: cin}
    }

    #[allow(dead_code)]
    fn read1<T: FromStr>(&mut self) -> Option<T> {
        let token = self.cin.by_ref().bytes().map(|c| c.unwrap() as char)
                    .skip_while(|c| c.is_whitespace())
                    .take_while(|c| !c.is_whitespace())
                    .collect::<String>();
        token.parse::<T>().ok()
    }

    #[allow(dead_code)]
    fn read<T: FromStr>(&mut self) -> T {
        self.read1().unwrap()
    }
}

fn main() {
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin);
    let n: usize = sc.read();
    let mut c: Vec<i32> = vec![0; n - 1];
    let mut s: Vec<i32> = vec![0; n - 1];
    let mut f: Vec<i32> = vec![0; n - 1];
    for i in 0..n-1 {
        c[i] = sc.read();
        s[i] = sc.read();
        f[i] = sc.read();
    }

    let mut ans: Vec<i32> = vec![INF; n];
    ans[n - 1] = 0;
    for i in 0..n-1 {
        let mut t = 0;
        for j in i..n-1 {
            t = max(t, s[j]);
            if (t - s[j]) % f[j] != 0 {
                t += f[j] - ((t - s[j]) % f[j]);
            }
            t += c[j];
        }
        ans[i] = t;
    }

    for x in ans {
        println!("{}", x);
    }
}