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>
 
#include <iostream>
#include <algorithm>
 
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <stack>
#include <queue>
#include <deque>
#include <cstring>
#include <string>
#include <utility>
#include <array>
#include <complex>
#include <valarray>
 
#include <cassert>
#include <cmath>
#include <functional>
#include <iomanip>
#include <chrono>
#include <random>
#include <numeric>
 
 
using namespace std;
#define int long long
 
typedef long long ll;
typedef unsigned long long ull;
//typedef unsigned __int128 HASH;
typedef pair<int,int> pii;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pullull;
typedef pair<ll,int> plli;
typedef pair<double, int> pdbi;
typedef pair<int,pii> pipii;
typedef pair<ll,pll> plpll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<pii> vpii;
typedef vector<vector<int>> mat;
 
#define rep(i,n) for (int i=0;i<(n);i++)
#define rep2(i,a,b) for (int i=(a);i<(b);i++)
#define rrep(i,n) for (int i=(n);i>0;i--)
#define rrep2(i,a,b) for (int i=(a);i>b;i--)
#define pb push_back
#define fi first
#define se second
#define all(a) (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
 
const ll hmod1 = 999999937;
const ll hmod2 = 1000000000 + 9;
const int INF = 1<<30;
const ll INFLL = 1LL<<62;
const double EPS = 1e-12;
const ll mod = 1000000000 + 7;
const int dx4[4] = {1, 0, -1, 0};
const int dy4[4] = {0, 1, 0, -1};
const int dx8[8] = {1, 1, 1, 0, 0, -1, -1, -1};
const int dy8[8] = {0, 1, -1, 1, -1, 0, 1, -1};
const double pi = 3.141592653589793;
 
#define addm(X, Y) (X) = ((X) + ((Y) % mod) + mod) % mod
#define inside(y, x, h, w) (0 <= (y) && (y) < (h) && 0 <= (x) && (x) < (w)) ? true : false

//debug
#define DEBUG 
 
#define DUMPOUT cout
 
#ifdef DEBUG
#define dump(...) DUMPOUT<<#__VA_ARGS__<<" :["<<__FUNCTION__<<":"<<__LINE__<<"]"<<endl; DUMPOUT<<"    "; dump_func(__VA_ARGS__)
#else
#define dump(...) 
#endif

void dump_func() {DUMPOUT << endl;};

template <class Head, class... Tail> void dump_func(Head&& head, Tail&&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) == 0) DUMPOUT << " ";
    else DUMPOUT << ", ";
    dump_func(std::move(tail)...);
}

//ostream 
template<typename T> ostream& operator << (ostream& os, vector<T>& vec) {
    os << "["; for (int i = 0; i<vec.size(); i++) os << vec[i] << (i + 1 == vec.size() ? "" : ", "); os << "]";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, pair<T, U>& pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}

template<typename T, typename U> ostream& operator << (ostream& os, map<T, U>& map_var) {
    os << "[";
    for (auto itr = map_var.begin(); itr != map_var.end(); itr++) {
        os << "(" << itr->first << ", " << itr->second << ")"; itr++;  if(itr != map_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}

template<typename T> ostream& operator << (ostream& os, set<T>& set_var) {
    os << "[";
    for (auto itr = set_var.begin(); itr != set_var.end(); itr++) {
        os << *itr; ++itr; if(itr != set_var.end()) os << ", "; itr--;
    }
    os << "]";
    return os;
}


int n;
int c[505], s[505], f[505];

signed main() {	
    cin.tie(0);
    ios::sync_with_stdio(false);
	cin >> n;
	rep(i, n - 1) cin >> c[i] >> s[i] >> f[i];
	rep(i, n - 1) {
		int now = i + 1;
		int x = s[i] + c[i];
		while (now < n - 1) {
			if (s[now] > x) x = s[now];
			//cout << x - s[now] << endl;
			if ((x - s[now]) % f[now] == 0) x += c[now];
			else {
				x = s[now] + (x - s[now]) / f[now] * f[now] + f[now];
				x += c[now]; 
			}
			now++;
		}
		cout << x << endl;
	}
	cout << 0 << endl;
}