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)

#define rep(i, a, b) for(int i = a; i < (int)(b); i++)
#include <bits/stdc++.h>

typedef long long ll;
using namespace std;
template<class T> inline void chmax(T &a, const T &b) {if(a < b) a = b;}
template<class T> inline void chmin(T &a, const T &b) {if(a > b) a = b;}
const long long INF = 1LL<<60;
const long long MOD = (ll)1e9+7;
const long long MAX = (ll)1e5+1000;

class stat{
   public:
   ll c;
   ll s;
   ll f;
   stat(ll a,ll a1,ll a2) {c=a;s =a1;f=a2;};
};
int main(void) {
   ll N;
   cin>>N;
   vector<stat*> vS(N-1);
   rep(i,0,N-1) {
      ll c,s,f;
      cin>>c>>s>>f;
      vS[i] = new stat(c,s,f);
   }
   rep(i,0,N) {
      ll now = 0;
      rep(j,i,N-1) {
         if (now < vS[j]->s) now = vS[j]->s;
         else now = (1 + (int)(now-1)/vS[j]->f) * vS[j]->f;
         now += vS[j]->c;
         // cout<<i<<" "<<j<<" "<<now<<endl;
      }
      cout<<now<<endl;
   }



   return 0;
}