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(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let N: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let (C, S, F): (Vec<i64>, Vec<i64>, Vec<i64>) = {
        let (mut C, mut S, mut F) = (vec![], vec![], vec![]);
        for _ in 0..(N - 1) {
            let mut line: String = String::new();
            std::io::stdin().read_line(&mut line).unwrap();
            let mut iter = line.split_whitespace();
            C.push(iter.next().unwrap().parse().unwrap());
            S.push(iter.next().unwrap().parse().unwrap());
            F.push(iter.next().unwrap().parse().unwrap());
        }
        (C, S, F)
    };

    fn bis<P: Fn(i64) -> bool>(ok: i64, ng: i64, p: P) -> i64 {
        let mid = (ok + ng) / 2;
        if (ok - ng).abs() == 1 { ok }
        else if p(mid) { bis(mid, ng, p) }
        else { bis(ok, mid, p) }
    }
    const INF: i64 = 1e15 as i64;

    let ans = (0..N).map(|i| {
        if i == N - 1 {
            0.to_string()
        } else {
            let mut x = 0;
            for j in i..(N - 1) {
                x = S[j] + bis(INF, -1, |k| x <= S[j] + k * F[j]) * F[j];
                // println!("j={}, k={}, x={}", j, bis(INF, -1, |k| x <= S[j] + k * F[j]), x);
                x += C[j];
            }
            x.to_string()
        }
    }).collect::<Vec<String>>().join("\n");

    println!("{}", ans);
}