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: Wrong Answer)

#include <iostream>
#include <vector>
#include<list>
#include <algorithm>
#include <string>
#include <math.h>
#include<map>
#include <utility>

#include<queue>
using namespace std;
#define REP(i,k,n) for(int i=k;i<n;i++)
 
#define INF 1844674407370955161
typedef long long int ll;
#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))
#define MOD 1000000007

template<typename T>
int vector_finder(std::vector<T> vec, T number) {
  auto itr = std::find(vec.begin(), vec.end(), number);
  size_t index = std::distance( vec.begin(), itr );
  if (index != vec.size()) { // 発見できたとき
    return 1;
  }
  else { // 発見できなかったとき
    return 0;
  }
}

template<typename T>
T gcd(T a,T b){
	if(b>0){
		return gcd(b, a % b);
	}else{
		return a;
	}
}

template<typename T>
T lcm(T a,T b){
	T g = gcd(a, b);
	return a /g* b ;
}

void clear( std::queue<pair<int,int> > &q )
{
   std::queue<pair<int,int> > empty;
   std::swap( q, empty );
}
bool IsPrime(int num)
{
    if (num < 2) return false;
    else if (num == 2) return true;
    else if (num % 2 == 0) return false; // 偶数はあらかじめ除く

    double sqrtNum = sqrt(num);
    for (int i = 3; i <= sqrtNum; i += 2)
    {
        if (num % i == 0)
        {
            // 素数ではない
            return false;
        }
    }

    // 素数である
    return true;
}
template<typename T>
int  GetDigit(T num){
    int digit=0;
    while(num!=0){
        num /= 10;
        digit++;
    }
    return digit;
}

template<typename T>
std::vector<T> enum_div(T n)//nの約数を列挙
{
	std::vector<T> ret;
	for(T i=1 ; i*i<=n ; ++i)
	{
		if(n%i == 0)
		{
			ret.push_back(i);
			if(i!=1 && i*i!=n)
			{
				ret.push_back(n/i);
			}
		}
	}
	return ret;
}


class DisjointSet{
    public:
        vector<int> rank, p;

        DisjointSet(){
        }

        DisjointSet(int size){
            rank.resize(size, 0);
            p.resize(size, 0);
            for (int i = 0; i < size;i++){
                makeSet(i);
            }
        }
        void makeSet(int x){
            p[x] = x;
            rank[x] = 0;
        }

        bool same(int x,int y){
            return findSet(x) == findSet(y);
        }
        void unite(int x,int y){
            link(findSet(x), findSet(y));
        }
        void link(int x,int y){
            if(rank[x]>rank[y]){
                p[y] = x;
            }else{
                p[x] = y;
                if(rank[x]==rank[y]){
                    rank[y]++;
                }
            }
        }
        int findSet(int x){
            if(x!=p[x]){
                p[x] = findSet(p[x]);
            }
            return p[x];
        }
};



int main()
{
    
    int N;
    int C[N],S[N],F[N];
    REP(i,0,N){
        cin>>C[i]>>S[i]>>F[i];
    }
    for ( int i =0; i < N ; i ++)
    {
    int 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 = t + F [ j ] - t % F [ j ];
    t += C [ j ];
    }
    cout<<t<<endl;
    }

}