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;
typedef long long ll;
const double EPS = 1e-9;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LINF = 1e18;
typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef pair<int, int> pi;
typedef pair<ll, ll> pll;
typedef map<int, int> mi;
typedef set<int> si;
#define VV(T) vector<vector< T > >
#define dump(x)  cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define POSSIBLE(n) cout << ((n) ? "POSSIBLE" : "IMPOSSIBLE"  ) << endl
#define Possible(n) cout << ((n) ? "Possible" : "Impossible"  ) << endl

#define rep(i, n) REP(i, 0, n)                              // 0, 1, ..., n-1
#define REP(i, x, n) for(int i = x; i < n; i++)             // x, x + 1, ..., n-1
#define FOREACH(x,a) for(auto& (x) : (a) )

#define ALL(v) (v).begin() , (v).end()
#define RALL(v) (v).rbegin(), (v).rend()

#define pb push_back
#define pu push
#define mp make_pair
#define fi first
#define sc second

#define COUT(x) cout << (x) << endl
#define VECCIN(x) for(auto&youso_: (x) )cin>>youso_
#define VECCOUT(x) for(auto&youso_: (x) )cout<<youso_<<" ";cout<<endl

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }


int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N; cin >> N;
    vi C(N), S(N), F(N);
    rep (i, N - 1) cin >> C[i] >> S[i] >> F[i];

    rep (i, N) {
        int total = 0;
        REP(j, i, N - 1) {
            total = max(total, S[j]);
            if (total % F[j] == 0) total += C[j];
            else total = total + F[j] - total % F[j] + C[j];
        }
        COUT(total);
    }
}