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 5.4.1) (judge verdict: Wrong Answer)

#define _CRT_SECURE_NO_WARNINGS

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>

/* define */
#define Yes(X) puts ( X ? "Yes" : "No")
#define YES(X) puts ( X ? "YES" : "NO")
#define FOR(i, a, b) for ( int i = a; i < b ; i++)
#define REP(i, n) for ( int i = 0; i < n ; i++)
#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define max3(a, b, c) ((max((a), (b)) > (c)) ? max((a), (b)) : (c))
#define min3(a, b, c) ((min((a), (b)) < (c)) ? min((a), (b)) : (c))

/* const */
const int MOD = (int)1e9 + 7;

/* alias */
typedef long long ll;
typedef unsigned  u;

/* function */
int cmp(const void *a, const void *b) {
	return *(int*)a - *(int*)b;
}
int cmp2(const void *a, const void *b) {
	return *(int*)b - *(int*)a;
}
int gcd(int a, int b) {
	if (b != 0)return gcd(b, a % b);
	return a;
}
ll Lgcd(ll a, ll b) {
	if (b != 0) return Lgcd(b, a % b); 
	return a;
}
int lcm(int a, int b) {
	return a / gcd (a, b) * b;
}
ll Llcm(ll a, ll b) {
	return a / Lgcd(a, b) * b;
}

/* main */
int main() {
	int n, t, c[500], s[500], f[500];
	scanf("%d", &n);
	REP(i, n - 1)scanf("%d%d%d", &c[i], &s[i], &f[i]);
	REP(i, n - 1) {
		t = s[i] + c[i];
		FOR(j, i, n - 2) {
			if (t <= s[j + 1])t = s[j + 1];
			else t = s[j + 1] + ((t - s[j + 1])/ f[j + 1]) * f[j + 1];
			t += c[j + 1];
		}
		printf("%d\n", t);
	}
	puts("0");
	return 0;
}