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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
#define ll int64_t
#define _GLIBCXX_DEBUG
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
//拡張ユークリッドの互除法
template< typename T >
T extgcd(T a, T b, T &x, T &y) {
  T d = a;
  if(b != 0) {
    d = extgcd(b, a % b, y, x);
    y -= (a / b) * x;
  } else {
    x = 1;
    y = 0;
  }
  return d;
}

//約数列挙(√N)
vector< int64_t > divisor(int64_t n) {
  vector< int64_t > ret;
  for(int64_t i = 1; i * i <= n; i++) {
    if(n % i == 0) {
      ret.push_back(i);
      if(i * i != n) ret.push_back(n / i);
    }
  }
  sort(begin(ret), end(ret));
  return (ret);
}
//素因数分解(√N)
map< int64_t, int > prime_factor(int64_t n) {
  map< int64_t, int > ret;
  for(int64_t i = 2; i * i <= n; i++) {
    while(n % i == 0) {
      ret[i]++;
      n /= i;
    }
  }
  if(n != 1) ret[n] = 1;
  return ret;
}
//素数テーブル(NloglogN)
vector< bool > prime_table(int n) {
  vector< bool > prime(n + 1, true);
  if(n >= 0) prime[0] = false;
  if(n >= 1) prime[1] = false;
  for(int i = 2; i * i <= n; i++) {
    if(!prime[i]) continue;
    for(int j = i + i; j <= n; j += i) {
      prime[j] = false;
    }
  }
  return prime;
}
//二項係数(K)
template< typename T >
T binomial(int64_t N, int64_t K) {
  if(K < 0 || N < K) return 0;
  T ret = 1;
  for(T i = 1; i <= K; ++i) {
    ret *= N--;
    ret /= i;
  }
  return ret;
}
//二項係数テーブル(N^2)
template< typename T >
vector< vector< T > > binomial_table(int N) {
  vector< vector< T > > mat(N + 1, vector< T >(N + 1));
  for(int i = 0; i <= N; i++) {
    for(int j = 0; j <= i; j++) {
      if(j == 0 || j == i) mat[i][j] = 1;
      else mat[i][j] = mat[i - 1][j - 1] + mat[i - 1][j];
    }
  }
  return mat;
}
//Gragh template
template< typename T >
struct edge {
  int src, to;
  T cost;

  edge(int to, T cost) : src(-1), to(to), cost(cost) {}

  edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

  edge &operator=(const int &x) {
    to = x;
    return *this;
  }

  operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WeightedGraph = vector< Edges< T > >;
using UnWeightedGraph = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;
//unionfind
struct UnionFind {
  vector< int > data;
 
  UnionFind(int sz) {
    data.assign(sz, -1);
  }
 
  bool unite(int x, int y) {
    x = find(x), y = find(y);
    if(x == y) return (false);
    if(data[x] > data[y]) swap(x, y);
    data[x] += data[y];
    data[y] = x;
    return (true);
  }
 
  int find(int k) {
    if(data[k] < 0) return (k);
    return (data[k] = find(data[k]));
  }
 
  int size(int k) {
    return (-data[find(k)]);
  }
};
//DIJKSTRA(ElogV)最短路
template< typename T >
vector< T > dijkstra(WeightedGraph< T > &g, int s) {
  const auto INF = numeric_limits< T >::max();
  vector< T > dist(g.size(), INF);

  using Pi = pair< T, int >;
  priority_queue< Pi, vector< Pi >, greater< Pi > > que;
  dist[s] = 0;
  que.emplace(dist[s], s);
  while(!que.empty()) {
    T cost;
    int idx;
    tie(cost, idx) = que.top();
    que.pop();
    if(dist[idx] < cost) continue;
    for(auto &e : g[idx]) {
      auto next_cost = cost + e.cost;
      if(dist[e.to] <= next_cost) continue;
      dist[e.to] = next_cost;
      que.emplace(dist[e.to], e.to);
    }
  }
  return dist;
}
//prim最小全域木
template< typename T >
T prim(WeightedGraph< T > &g) {
  using Pi = pair< T, int >;
 
  T total = 0;
  vector< bool > used(g.size(), false);
  priority_queue< Pi, vector< Pi >, greater< Pi > > que;
  que.emplace(0, 0);
  while(!que.empty()) {
    auto p = que.top();
    que.pop();
    if(used[p.second]) continue;
    used[p.second] = true;
    total += p.first;
    for(auto &e : g[p.second]) {
      que.emplace(e.cost, e.to);
    }
  }
  return total;
}
//bellman_ford(VE)単一始点最短路
template< typename T >
vector< T > bellman_ford(Edges< T > &edges, int V, int s) {
  const auto INF = numeric_limits< T >::max();
  vector< T > dist(V, INF);
  dist[s] = 0;
  for(int i = 0; i < V - 1; i++) {
    for(auto &e : edges) {
      if(dist[e.src] == INF) continue;
      dist[e.to] = min(dist[e.to], dist[e.src] + e.cost);
    }
  }
  for(auto &e : edges) {
    if(dist[e.src] == INF) continue;
    if(dist[e.src] + e.cost < dist[e.to]) return vector< T >();
  }
  return dist;
}


//memo cout << fixed << setprecision(桁数);//
int main(){
  int n;
  cin >> n;
  vector<int> a(n);
  rep(i,n-1){
    int t,s,f;
    cin >> t>>s>>f;
    rep(j,i+1){
      int now=max(a.at(j),s);
      a.at(j)=(now/f)*f;
      a.at(j)+=t;
    }
  }
  rep(i,n-1)
    cout << a.at(i) << endl;
  cout << 0 << endl;
}