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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<list>
#include<set>
#include<map>
#include<utility>
#include<cmath>


#define rep(i,x) for(int i=0;i<(int)(x);i++)
#define reps(i,x) for(int i=1;i<=(int)(x);i++)
#define rrep(i,x) for(int i=((int)(x)-1);i>=0;i--)
#define rreps(i,x) for(int i=((int)(x));i>0;i--)

#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define INF 2e9

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
typedef long long ll;
typedef std::pair<ll, ll> P;
int gcd(int a,int b){return b?gcd(b,a%b):a;}

using namespace std;

int main()
{
    ll n;
    cin >> n;
    vector<ll> c(n-1), s(n-1), f(n-1);

    rep(i, n-1){
        cin >> c[i] >> s[i] >> f[i];
    }

    rep(i, n){
        ll t = 0;
        for(int j=i; j<n-1; j++){
            if(t < s[j]) t = s[j];
            else if (t%f[j] == 0);
            else t += f[j] - t%f[j];
            t += c[j];
        }
        cout << t << endl;
    }
}