Codeforces Round #436 (Div. 2)-E-Fire（01背包输出路径）

E. Fire

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp is in really serious trouble — his house is on fire! It's time to save the most valuable items. Polycarp estimated that it would take ti seconds to save i-th item. In addition, for each item, he estimated the value of di — the moment after which the item i will be completely burned and will no longer be valuable for him at all. In particular, if ti ≥ di, then i-th item cannot be saved.

Given the values pi for each of the items, find a set of items that Polycarp can save such that the total value of this items is maximum possible. Polycarp saves the items one after another. For example, if he takes item a first, and then item b, then the item a will be saved in ta seconds, and the item b — in ta + tb seconds after fire started.

Input

The first line contains a single integer n (1 ≤ n ≤ 100) — the number of items in Polycarp's house.

Each of the following n lines contains three integers ti, di, pi (1 ≤ ti ≤ 20, 1 ≤ di ≤ 2 000, 1 ≤ pi ≤ 20) — the time needed to save the item i, the time after which the item i will burn completely and the value of item i.

Output

In the first line print the maximum possible total value of the set of saved items. In the second line print one integer m — the number of items in the desired set. In the third line print m distinct integers — numbers of the saved items in the order Polycarp saves them. Items are 1-indexed in the same order in which they appear in the input. If there are several answers, print any of them.

Examples

input

33 7 42 6 53 7 6

output

1122 3 

input

25 6 13 3 5

output

111 

Note

In the first example Polycarp will have time to save any two items, but in order to maximize the total value of the saved items, he must save the second and the third item. For example, he can firstly save the third item in 3 seconds, and then save the second item in another 2 seconds. Thus, the total value of the saved items will be 6 + 5 = 11.

In the second example Polycarp can save only the first item, since even if he immediately starts saving the second item, he can save it in 3 seconds, but this item will already be completely burned by this time.

题意：给你n个物品，第i个物品在d[i]秒爆炸，每次救这个物品需要花t[i]秒。

救下这个物品可获得的最大价值为p[i]，问你能获得的最大价值。

题解：01背包模板题，唯一区别就是输出路径，还是模板题。

#include<set>  #include<map>     #include<stack>            #include<queue>            #include<vector>    #include<string> #include<time.h>#include<math.h>            #include<stdio.h>            #include<iostream>            #include<string.h>            #include<stdlib.h>    #include<algorithm>   #include<functional>    using namespace std;            #define ll long long       #define inf  1000000000       #define mod 1000000007             #define maxn  2005#define lowbit(x) (x&-x)            #define eps 1e-9struct node{int t,d,p,id;}a[105];int ans[maxn],p[105][maxn],dp[105][maxn];bool comp(node a,node b){return a.d<b.d;}void solve(int x,int y,int sum){if(x<0)printf("%d\n",sum);else{if(!p[x][y])solve(x-1,y,sum);else{solve(x-1,y-a[x].t,sum+1);printf("%d ",a[x].id);}}}int main(void){int n,i,j,k;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d%d%d",&a[i].t,&a[i].d,&a[i].p);a[i].id=i;}sort(a+1,a+n+1,comp);for(i=1;i<=n;i++)for(j=0;j<=2000;j++){dp[i][j]=dp[i-1][j];if(j>=a[i].t && j<a[i].d)if(dp[i-1][j-a[i].t]+a[i].p>dp[i][j])dp[i][j]=dp[i-1][j-a[i].t]+a[i].p,p[i][j]=1;}int sum=-1,num=-1;for(i=0;i<=2000;i++)if(dp[n][i]>sum)sum=dp[n][i],num=i;printf("%d\n",sum);solve(n,num,0);return 0;}