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: Wrong Answer)

/*
ID:wmj20041
TASK:
LANG:C++
*/
#include<functional>
#include<algorithm>
#include<exception>
#include<stdexcept>
#include<streambuf>
#include<iterator>
#include<string.h>
#include<stdlib.h>
#include<typeinfo>
#include<valarray>
#include<iostream>
#include<sstream>
#include<istream>
#include<stdio.h>
#include<climits>
#include<clocale>
#include<complex>
#include<csetjmp>
#include<csignal>
#include<cstdarg>
#include<cstddef>
#include<ctype.h>
#include<cassert>
#include<cstdlib>
#include<utility>
#include<cstring>
#include<numeric>
#include<ostream>
#include<cwctype>
#include<fstream>
#include<iomanip>
#include<math.h>
#include<bitset>
#include<cctype>
#include<string>
#include<vector>
#include<limits>
#include<locale>
#include<memory>
#include<cerrno>
#include<time.h>
#include<iosfwd>
#include<cfloat>
#include<cstdio>
#include<cwchar>
#include<cmath>
#include<ctime>
#include<deque>
#include<queue>
#include<stack>
#include<list>
#include<ios>
#include<map>
#include<set>
#include<new>
#define LL long long
#define pb push_back
#define pob pop_back
#define mp make_pair
#define FILL(a,v) memset((void*)a,v,sizeof(a))
#define FILL0(a) FILL(a,0)
#define FILL1(a) FILL(a,-1)
#define FILLBIG(a) FILL(a,0x7f)
#define ALL(a) (a).begin(),(a).end()
#define Inf ((int)0x7f7f7f7f)
#define Mod 1000000007
#define foreach(it,x) for(__typeof(x.begin()) it=x.begin();it!=x.end();it++)
#define FOR(i,x) for(int i=0;i<x;i++)
#define RFOR(i,x) for(int i=x-1;i>=0;i--)
#define FFOR(i,a,b) for(int i=a;i<b;i++)
#define FRFOR(i,a,b) for(int i=a-1;i>=b;i--)
#define SZ(a) ((int)(a).size())
#define FI first
#define SE second
#define OVER(a) return puts(a),0
#define fastio ios::sync_with_stdio(0)
#define filei(f) freopen(f,"r",stdin)
#define fileo(f) freopen(f,"w",stdout)
#define eprintf(...) fprintf(stderr,__VA_ARGS__)
#define MASK(a) (1<<(a))
#define MASKX(x,a) ((x)<<(a))
#define BIT(a,b) ((a>>b)&1)
#define cddx(a) priority_queue<a >
#define cxdd(a) priority_queue<a,vector<a >,greater<a > >
#define deval(a) {cerr<<#a<<" = "<<a<<endl;}
#define dearr(be,en) {cerr<<#be;cerr<<" = {";for(auto i=be;i!=en;i++)cerr<<' '<<*i;cerr<<" }\n";}
#define reg register
using namespace std;
typedef pair<int,int> pii;
typedef unsigned long long UINT;
typedef unsigned int uint;
typedef long long INT;
typedef long double ldouble;
typedef pair<INT,INT> PII;
typedef stringstream sst;
typedef istringstream iss;
typedef vector<int> vint;
typedef vector<INT> VINT;
typedef vector<pii> vpii;
typedef vector<PII> VPII;
typedef priority_queue<int> pqi;
typedef priority_queue<INT> PQI;
typedef priority_queue<pii> pqp;
typedef priority_queue<PII> PQP;
typedef priority_queue<int,vint,greater<int> > xqi;
typedef priority_queue<INT,VINT,greater<INT> > XQI;
typedef priority_queue<pii,vpii,greater<pii> > xqp;
typedef priority_queue<PII,VPII,greater<PII> > XQP;
const int INF=1e9+7;
using namespace std;
int n;
long long ans,t[505],s[505],f[505];
int main()
{
	//freopen("input.in","r",stdin);
	//freopen("output.out","w",stdout);
	cin>>n;
	for (int i=1;i<n;i++)
	cin>>t[i]>>s[i]>>f[i];
	int i=1;
	for (int j=1;j<=n;j++)
	{
		i=j;
		ans=0;
	    while(i<n)
	    {
		    if (ans>=s[i]) 
    		{
    			int k;
    			for (k=s[i];k<=ans;k+=f[i]) ;
    			//cout<<k<<endl;
    			ans=k-f[i];
	    		ans+=t[i];
		    	i++;
			    continue;
    		}
			ans++;
	    }
	    cout<<ans<<endl;
	}
	return 0;
}