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<bits/stdc++.h>
using namespace std;

#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(a) (int)(a.size())
#define ALL(a) a.begin(),a.end()

#define SI(n) scanf("%d",&n)
#define PI(n) printf("%d\n",n)
#define SL(n) scanf("%lld",&n)
#define PL(n) printf("%lld\n",n)
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)

#define TRACE
#ifdef TRACE
#define TR(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');cerr.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
#else
#define TR(...)
#endif
//FILE *fin = freopen("in","r",stdin);
//FILE *fout = freopen("out","w",stdout);
typedef long long int 	LL;
typedef long double 	LD;
typedef pair<int,int>   II;
typedef vector<int>     VI;
const int MOD = 1e9 + 7;
const int N = 100011;
int C[N], F[N], S[N];

int main() {
    int n;
    SI(n);
    for(int i = 0; i < n - 1; ++i) {
        SI(C[i]), SI(S[i]), SI(F[i]);
    }
    
    for(int i = 0; i < n - 1; ++i) {
        int time = S[i] + C[i];
        for(int j = i + 1; j < n - 1; ++j) {
            if(time < S[j])
                time = S[j];
            time += time % F[j] ? (F[j] - time % F[j]) : 0;
            time += C[j];
        }
        PI(time);
    }
    PI(0);

    return 0;
}