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: Wrong Answer)

fn main() {
    let n = read();

    // 0:C 1:S 2:F
    let vec = read_lines(n - 1);
    let mut ans = Vec::new();
    for j in 0..n - 1 {
        let mut cnt = vec[j][0] + vec[j][1];
        for i in j + 1..n - 1 {
            let c = vec[i][0];
            let s = vec[i][1];
            let f = vec[i][2];
            if cnt <= s {
                cnt = s + c;
            } else {
                // s + f * n
                // ((s/f) + n) * f
                let left = cnt - s;
                let step = (left as f64 / f as f64).floor() as i32;
                let right = f * step + s + c;
                cnt = right;
            }
        }
        ans.push(cnt);
    }

    for a in ans {
        println!("{}", a);
    }
    println!("0");
}

fn read() -> usize {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_lines(n: usize) -> Vec<Vec<i32>> {
    let mut v2 = Vec::new();
    for _ in 0..n {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).ok();
        let v = s.trim()
            .split_whitespace()
            .map(|e| e.parse().ok().unwrap())
            .collect();
        v2.push(v);
    }
    v2
}