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"
#include "math.h"
 
using namespace std;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<int> vin;
typedef pair<ll,ll> P;
typedef vector<P> vp;
#define rep(i,a,b) for(ll i=(a);i<(b);++i)
#define drep(i,a,b) for(ll i=(a);i>=(b);--i)
#define SIZE(a) int((a).size())
#define out(a) cout<<(a)<<endl;
const int INF=INT_MAX;
const int MAX = 510000;
const ll MOD = 1000000007;

int main(){
  ll n;cin>>n;
  vll c(n-1),s(n-1),f(n-1);
  rep(i,0,n-1){
    cin>>c[i]>>s[i]>>f[i];
  }
  rep(i,0,n-1){
    ll now=s[i];
    rep(j,i,n-1){
      ll tmp=(now-s[j])/f[j];
      if((now-s[j])%f[j]==0){
        now=s[j]+f[j]*tmp+c[j];
      }
      else if(now-s[j]<0){
        now=s[j]+c[j];
      }
      else now=s[j]+f[j]*(tmp+1)+c[j];
      
    }
    cout<<now<<endl;
  }
  cout<<0<<endl;
}