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++ (GCC 9.2.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
#define rep2(i, l, r) for (int i = (l); i < (r); i++)
#define ALL(x) (x).begin(), (x).end()     //昇順
#define RALL(x) (x).rbegin(), (x).rend()  // 降順
#define pri(x) cout << (x) << "\n"
#define pri2(x, y) cout << (x) << " " << (y) << "\n"
#define pri3(x, y, z) cout << (x) << " " << (y) << " " << (z) << "\n"
const long long mod = 1e9 + 7;
typedef long long ll;
typedef priority_queue<int> PQ;
typedef vector<long long> VL; // VL a(n);
typedef vector<bool> VB;
typedef vector<double> VD;
typedef vector<string> VS;
typedef vector<char> VC;
typedef vector<VS> VSS;
typedef vector<VC> VCC;
typedef vector<VL> VLL; // VII a(n,VI(m)) n * m
typedef pair<ll, ll> PL;
typedef map<ll, ll> MP; // MP a;
typedef vector<pair<ll, ll>> PS;

template <class T, class U>
bool chmax(T &a, U b) {
  if (a <= b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T, class U>
bool chmin(T &a, U b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "{";
  rep(i, (int)v.size()) { os << v[i] << (i < v.size() - 1 ? ", " : ""); }
  os << "}";
  return os;
}

// g++ -std=c++11 prac.cpp
// operator << (cout,a);
// chmin(min,a)

int main() {
  ll n,k,m,x=0,y=0,z = 0,h=0,w=0,ans=0,sum = 0, Max = -1, Min = 3e9+1;
  string t;
  bool ok=true; 
  cin >> n;
  VL c(n),s(n),f(n);
  rep(i,n-1){
    cin >> c[i] >> s[i] >> f[i];
  }
  rep(i,n-1){
    ans = 0;
    ans = s[i] + c[i];
    for(int j=i+1; j<n-1; j++){
      if(ans < s[j]){
        ans = s[j];
      }
      else if(ans % f[j] == 0){
      }
      else{
        ans = ans+f[j] - ans%f[j];
      }
      ans += c[j];
    }
    pri(ans);
  }
  pri(0);
  return 0;
}