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本文描述了一个关于孩子们在牙医诊所排队的故事。每个孩子都有不同的信心值，进入诊所看牙时会哭泣，并影响后续孩子的信心。若信心降至负数，则孩子会离开队伍。文章探讨了在一系列哭泣和离开后，哪些孩子最终能够接受治疗。

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Gennady is one of the best child dentists in Berland. Today n children got an appointment with him, they lined up in front of his office.

All children love to cry loudly at the reception at the dentist. We enumerate the children with integers from 1 to n in the order they go in the line. Every child is associated with the value of his cofidence p_i. The children take turns one after another to come into the office; each time the child that is the first in the line goes to the doctor.

While Gennady treats the teeth of the i-th child, the child is crying with the volume of v_i. At that the confidence of the first child in the line is reduced by the amount of v_i, the second one — by value v_i - 1, and so on. The children in the queue after the v_i-th child almost do not hear the crying, so their confidenceremains unchanged.

If at any point in time the confidence of the j-th child is less than zero, he begins to cry with the volume of d_j and leaves the line, running towards the exit, without going to the doctor's office. At this the confidence of all the children after the j-th one in the line is reduced by the amount of d_j.

All these events occur immediately one after the other in some order. Some cries may lead to other cries, causing a chain reaction. Once in the hallway it is quiet, the child, who is first in the line, goes into the doctor's office.

Help Gennady the Dentist to determine the numbers of kids, whose teeth he will cure. Print their numbers in the chronological order.

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 4000) — the number of kids in the line.

Next n lines contain three integers each v_i, d_i, p_i (1 ≤ v_i, d_i, p_i ≤ 10⁶) — the volume of the cry in the doctor's office, the volume of the cry in the hall and the confidence of the i-th child.

Output

In the first line print number k — the number of children whose teeth Gennady will cure.

In the second line print k integers — the numbers of the children who will make it to the end of the line in the increasing order.

Example

Input

Output

2
1 3

Input

Output

4
1 2 4 5

Note

In the first example, Gennady first treats the teeth of the first child who will cry with volume 4. The confidences of the remaining children will get equal to - 2, 1, 3, 1, respectively. Thus, the second child also cries at the volume of 1 and run to the exit. The confidence of the remaining children will be equal to 0, 2, 0. Then the third child will go to the office, and cry with volume 5. The other children won't bear this, and with a loud cry they will run to the exit.

In the second sample, first the first child goes into the office, he will cry with volume 4. The confidence of the remaining children will be equal to 5, - 1, 6, 8. Thus, the third child will cry with the volume of 1 and run to the exit. The confidence of the remaining children will be equal to 5, 5, 7. After that, the second child goes to the office and cry with the volume of 5. The confidences of the remaining children will be equal to 0, 3. Then the fourth child will go into the office and cry with the volume of 2. Because of this the confidence of the fifth child will be 1, and he will go into the office last.

此题就是1到n的熊孩子去看牙医，每个人都有一个v,d,p的值，当p>=0的时候他就进去看牙医，然后就开始哭，他一哭剩余队列的人就吓死了，信心值就开始减小，但是哭声有限，越后面减少的越小，如果轮到的人信心值小于0，他就直接跑了，在他后面的人一看有人跑了，就直接吓尿了，后面所有人全部信心值全部减少di，求几个进去看了牙医，分别是哪几个

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
using namespace std;
const int maxn=1e5+5;
typedef long long ll;
#define INF 9999999
struct node{
ll v,d,p;
}e[5005];
int main()
{
ll n;
ll num=0;
ll sum[4005];
ll t=0;
scanf("%lld",&n);
for(int i=1;i<=n;i++)
{
scanf("%lld%lld%lld",&e[i].v,&e[i].d,&e[i].p);
}
for(int i=1;i<=n;i++)
{
if(e[i].p>=0)
{
num++;
sum[t++]=i;
ll cnt=0;
for(int j=i+1;j<=n;j++)
{
if(e[j].p>=0)
{
e[j].p-=e[i].v+cnt;
if(e[i].v>0)
e[i].v--;
if(e[j].p<0)
cnt+=e[j].d;
}
}
}
}
printf("%lld\n",num);
for(int i=0;i<t;i++)
printf("%lld ",sum[i]);
printf("\n");
}