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)

#pragma region header
#include <bits/stdc++.h>
#define int long long
#define ll long long
#define ld long double
#define vi vector<int>
#define vvi vector<vector<int>>
#define vvvi vector<vector<vector<int>>>
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define SORT(a) stable_sort(all(a))
#define RSORT(a) stable_sort(rall(a))
#define rev(a) reverse(all(a))
#define uniq(a) a.erase(unique(all(a)), end(a))
#define lambda(i) [=](auto i)
#define compare(i, j) [=](auto i, auto j)
#define min_of(a) *min_element(all(a))
#define max_of(a) *max_element(all(a))
#define sum(a) accumulate(all(a), 0)
#define count_of(a, i) count(all(a), i)
#define lower(a, i) (lower_bound(all(a), i) - begin(a)) // use member func for set
#define upper(a, i) (upper_bound(all(a), i) - begin(a))
#define binary(a, i) binary_search(all(a), i)
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define sz(a) (int)(a).size()
#define rep1(i, n) for (decltype(+n) i = 0; i < (n); i++)
#define rrep1(i, n) for (auto i = n - 1; i != static_cast<decltype(i)>(-1); i--)
#define rep2(i, a, b) for (auto i = (a); i < (b); i++)
#define rrep2(i, a, b) for (auto i = b - 1; i >= a; i--)
#define GET_MACRO(_1, _2, _3, NAME, ...) NAME
#define rep(...) GET_MACRO(__VA_ARGS__, rep2, rep1) (__VA_ARGS__)
#define rrep(...) GET_MACRO(__VA_ARGS__, rrep2, rrep1) (__VA_ARGS__)
#define each(i, a) for (auto &&i : (a))
using namespace std;

const int INF = 1e18;
const int MOD = 1e9 + 7;
int mod_add(int A, int B) { return (A + B) % MOD; }
int mod_sub(int A, int B) { return (A + MOD - B) % MOD; }
int mod_mul(int A, int B) { return ((A % MOD) * (B % MOD)) % MOD; }
int mod_bipow(int X, int Y) {
  if (Y == 0) return 1;
  else if (Y == 1) return X % MOD;
  else if (Y % 2 == 0) {
    int Z = mod_bipow(X, (int)(Y / 2));
    return mod_mul(Z, Z);
  } else {
    int Z = mod_bipow(X, (int)(Y / 2));
    return mod_mul(mod_mul(Z, Z), X);
  }
}
int mod_inv(int X) { return mod_bipow(X, MOD - 2); }
int mod_div(int A, int B) { return mod_mul(A, mod_inv(B)); }

template <class T>
bool amin(T &A, T B) {
  if (A > B) {
    A = B;
    return true;
  }
  return false;
}
template <class T>
bool amax(T &A, T B) {
  if (A < B) {
    A = B;
    return true;
  }
  return false;
}

vi divisors(int N) {
  vi ret;
  for (int i = 1; i * i <= N; i++) {
    if (N % i == 0) {
      ret.pb(i);
      if (i * i != N) ret.pb(N / i);
    }
  }
  SORT(ret);
  return ret;
}

map<int, int> prime_factors(int N) {
  map<int, int> ret;
  for (int i = 2; i * i <= N; i++) {
    while (N % i == 0) {
      ret[i]++;
      N /= i;
    }
  }
  if (N != 1) ret[N]++;
  return ret;
}

int ceil_div(int X, int Y) { return (X - 1) / Y + 1; }

vi cum(vi V) {
  vi U(sz(V));
  partial_sum(all(V), begin(U));
  return U;
}

struct union_find {
  vi data;
  union_find(int size) : data(size, -1) {}
  bool union_set(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
    return x != y;
  }
  bool find_set(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  int size(int x) { return -data[root(x)]; }
};

struct vertex {
  vector<int> adj = vi(0);
};
#pragma endregion header

// MOD = 1e9 + 7;


void solve(int n, vi c, vi s, vi f){
  vi t(n);
  rep(i, n){
    int t = 0;
    rep(j, i, n - 1){
      if (t < s[j]) t = s[j] + c[j];
      else t = s[j] + ceil_div(t - s[j], f[j]) * f[j] + c[j];
    }
    cout << t << endl;
  }
}


#pragma region main
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout << fixed << setprecision(15);

  int n;
  cin >> n;
  vi c(n-1);
  vi s(n-1);
  vi f(n-1);
  for(int i = 0 ; i < n-1 ; i++){
    cin >> c[i];
    cin >> s[i];
    cin >> f[i];
  }
  solve(n, move(c), move(s), move(f));
}
#pragma endregion main