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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.io.*;
import java.math.*;
 
 
import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;
 
public class Main{ 
 
 
    static long mod=1000000007;
    static int dx[]={1,-1,0,0};
    static int dy[]={0,0,1,-1};
    // static int dx[]={1,-1,0,0,1,1,-1,-1};
    // static int dy[]={0,0,1,-1,1,-1,1,-1};
    // PriorityQueue<Integer> que = new PriorityQueue<Integer>(); 
    //HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
 
    public  static void main(String[] args)   throws Exception, IOException{
     
        
        Reader sc = new Reader(System.in);
        PrintWriter out=new PrintWriter(System.out);
     
        int n=sc.nextInt();
        // long n=sc.nextLong();
        // char c[][] = new char[h][w];
        // long a[] = new long [n];
        // for (int i=0; i<n; i++) {a[i]=sc.nextInct();}
        // char c[] = sc.nextString().toCharArray();
        n--;
        int c[]=new int[n];
        int s[]=new int[n];
        int f[]=new int[n];
        for (int i=0; i<n; i++) {
            c[i]=sc.nextInt();
            s[i]=sc.nextInt();
            f[i]=sc.nextInt();
        }

        for (int i=0; i<n; i++) {
            long w=(long)s[i]+c[i];
            for (int t=i+1; t<n; t++) {
                int rm=(int)((w-s[t])%f[t]);
                if(s[t]>w)w=(long)s[t]+c[t];
                else w=w+(rm==0?0:f[t]-rm)+c[t];
            }
            out.println(w);
        }

        out.println(0);
        out.flush();
    }

    static void db(Object... os){
     
        System.err.println(Arrays.deepToString(os));
     
    }
     
    static boolean validpos(int x,int y,int r, int c){
        
        return x<r && 0<=x && y<c && 0<=y;
        
    }
     
    static boolean bit(long x,int k){
        // weather k-th bit (from right) be one or zero
        return  ( 0 < ( (x>>k) & 1 )  )  ? true:false;
    }    
}

class XY {
    int x,y,d;
    XY(int x, int y, int d) {
        this.x=x;
        this.y=y;
        this.d=d;
    } 
}
 
class P implements Comparable<P>{
    int x,y;
    P(int x, int y) {
        this.x=x;
        this.y=y;
    } 
      
    public int compareTo(P p){
        return -x + p.x;//des
    } 
}
 
class Reader
{ 
    private BufferedReader x;
    private StringTokenizer st;
    
    public Reader(InputStream in)
    {
        x = new BufferedReader(new InputStreamReader(in));
        st = null;
    }
    public String nextString() throws IOException
    {
        while( st==null || !st.hasMoreTokens() )
            st = new StringTokenizer(x.readLine());
        return st.nextToken();
    }
    public int nextInt() throws IOException
    {
        return Integer.parseInt(nextString());
    }
    public long nextLong() throws IOException
    {
        return Long.parseLong(nextString());
    }
    public double nextDouble() throws IOException
    {
        return Double.parseDouble(nextString());
    }
}