Problem: 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

3
3 7 4
2 6 5
3 7 6


Output

11
2
2 3 


Input

2
5 6 1
3 3 5


Output

1
1
1 

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.

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Main {
	static  ArrayList<Integer>adj[];
	static PrintWriter out=new PrintWriter(System.out);
	 public static int mod = 1000000007;

	static Boolean notmemo[][][];
	static int k;
	static Triple a[];
	static long b[];
	static int m;
	static int trace[];
	public static void main(String[] args) throws Exception {
	Scanner sc=new Scanner(System.in);
	 n=sc.nextInt();
	a=new Triple[n];
	for (int i = 0; i < a.length; i++) {
		a[i]=new Triple(sc.nextInt(), sc.nextInt(),sc.nextInt(),i);
	}
	Arrays.sort(a);
	memo=new long[n][4000];
	for (int i = 0; i < memo.length; i++) {
		Arrays.fill(memo[i], -1);
	}
	out.println(dp(0,0));
	trace(0,0);
	out.println(l1.size());
	Collections.reverse(l1);
	
	for(int x:l1) {
		out.print((x+1)+" ");
	}
	out.flush();
	
	
	out.flush();
	}
	static int teams=0;
static long mini[];
static long x[];
static long s[];
static ArrayList<Integer> l1=new ArrayList<>();
private static void trace(int i, int time) {
if(i==n) 
	return;


long ans=dp(i,time);
if(time+a[i].t<a[i].burn&&(ans==dp(i+1,time+a[i].t)+a[i].cost)) {
	
	trace(i+1, time+a[i].t);
    
	l1.add(a[i].idx);
    return;
}
    trace(i+1,time);

}


static long count=0;
	static long dp(int i,int time) {
    if(i==n) {
    	return 0;
    }
    
    if(memo[i][time]!=-1) {
    	return memo[i][time];
    }
    long ans=dp(i+1,time);
    
    
    if(time+a[i].t<a[i].burn)
    ans=Math.max(ans,dp(i+1,time+a[i].t)+a[i].cost);

	return memo[i][time]=ans;
	
	}
	

	static class incpair implements Comparable<incpair>
{
	int a; long b;
	int idx;
	incpair(int a, long dirg,int i) {this.a = a;	b = dirg; idx=i; }
	
	public int compareTo(incpair e){ return (int) (b - e.b);	}
}
	static class decpair implements Comparable<decpair>
{
	int a; long b;int idx;
	
	decpair(int a, long dirg,int i) {this.a = a;	b = dirg; idx=i;}
	
	public int compareTo(decpair e){ return (int) (e.b- b);	}
}

	
	
	
	
	
	


	static long allpowers[];
	static class Quad implements Comparable<Quad>{
		int u;
		int v;
		char state;
		int turns;
		public Quad(int i, int j, char c, int k) {
		u=i;
		v=j;
		state=c;
		turns=k;
		}
		public int compareTo(Quad e){ return (int) (turns - e.turns);	}
		
	}
	static long dirg[][];
	static Edge[] driver;
	
	static int n;
		static class Edge implements Comparable<Edge>
	{
		int node; long cost;
		
		Edge(int a, long dirg) { node = a;	cost = dirg; }
		
		public int compareTo(Edge e){ return (int) (cost - e.cost);	}
	}

	



	  static long manhatandistance(long x,long x2,long y,long y2) {
		  return  Math.abs(x-x2)+Math.abs(y-y2);
	  }
	  static long fib[];
	  static long fib(int n) {
		  if(n==1||n==0) {
			  return 1;
		  }
		  if(fib[n]!=-1) {
			  return fib[n];
		  }
		  else 
			  return  fib[n]=  ((fib(n-2)%mod+fib(n-1)%mod)%mod);
	  }
	  static class Pair implements Comparable<Pair>{
		  long val;
		  int idx;
		  Pair(int a, long b){
			  idx=a;
			  val=b;
		  }
		public int compareTo(Pair p) {
		
			return (int) (this.val-p.val);
		}
		  
		  
	
		
		
	}
	static long[][] comb;		
	static long nCr1(int n , int k)
	{
		if(n < k)
			return 0;
		if(k == 0 || k == n)
			return 1;
		if(comb[n][k] != -1)
			return comb[n][k];
		if(n - k < k)
			return comb[n][k] = nCr1(n, n - k);
		return comb[n][k] = (nCr1(n - 1, k - 1) + nCr1(n - 1, k))%mod;
	}
	
	
static  class Triple implements Comparable<Triple>{

int t;
int burn;
		int cost;
		int idx;
		public Triple(int a,int b,int l,int i) {
			t=a;
			burn=b;
			cost=l;
			idx=i;
		}
		@Override
		public int compareTo(Triple o) {
			return Long.compare( burn,o.burn);	
		}
		
	}
	
	static TreeSet<Long> primeFactors(long N)		// O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(p * p <= N)
		{
			while(N % p == 0) { factors.add((long) p); N /= p; }
			if(primes.size()>idx+1)
			p = primes.get(++idx);
			else
				break;
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}
	
	static boolean visited[];
	/**static int bfs(int s)
	{
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(s);
		int count=0;
		int maxcost=0;
		int dist[]=new int[n];
		dist[s]=0;
		while(!q.isEmpty())
		{
			
			int u = q.remove();
			if(dist[u]==k) {
				break;
			}
			for(Pair v: adj[u])
			{	
				maxcost=Math.max(maxcost, v.cost);
			
			
			
			if(!visited[v.v])	{
				
				visited[v.v]=true;
				q.add(v.v);
				dist[v.v]=dist[u]+1;
				maxcost=Math.max(maxcost, v.cost);
			}
			}	
			
		}
		return maxcost;
	}**/
	public static boolean FindAllElements(int n, int k) {
	 int sum = k;  
     int[] A = new int[k];  
      Arrays.fill(A, 0, k, 1);  
        
      for (int i = k - 1; i >= 0; --i) {  
      
           while (sum + A[i] <= n) {  
      
              sum += A[i];  
              A[i] *= 2;  
          }  
      }  
      if(sum==n) {
      	return true;
      }
      else
    	return false; 
	}
	
	static long[][] memo;

	
	static boolean vis2[][];
	
	static class SegmentTree {		// 1-based DS, OOP
		
		int N; 			//the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;
		
		SegmentTree(int[] in)		
		{
			array = in; N = in.length - 1;
			sTree = new int[N<<1];		//no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N<<1];
			build(1,1,N);
		}
		
		void build(int node, int b, int e)	// O(n)
		{
			if(b == e)					
				sTree[node] = array[b];
			else						
			{
				int mid = b + e >> 1;
				build(node<<1,b,mid);
				build(node<<1|1,mid+1,e);
				sTree[node] = sTree[node<<1]+sTree[node<<1|1];
			}
		}
		
		
		void update_point(int index, int val)			// O(log n)
		{
			index += N - 1;				
			sTree[index] += val;			
			while(index>1)				
			{
				index >>= 1;
				sTree[index] = sTree[index<<1] + sTree[index<<1|1];		
			}
		}
		
		
		void update_range(int i, int j, int val)		// O(log n) 
		{
			update_range(1,1,N,i,j,val);
		}
		
		void update_range(int node, int b, int e, int i, int j, int val)
		{
			if(i > e || j < b)		
				return;
			if(b >= i && e <= j)		
			{
				sTree[node] += (e-b+1)*val;			
				lazy[node] += val;				
			}							
			else		
			{
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node<<1,b,mid,i,j,val);
				update_range(node<<1|1,mid+1,e,i,j,val);
				sTree[node] = sTree[node<<1] + sTree[node<<1|1];		
			}
			
		}
		void propagate(int node, int b, int mid, int e)		
		{
			lazy[node<<1] += lazy[node];
			lazy[node<<1|1] += lazy[node];
			sTree[node<<1] += (mid-b+1)*lazy[node];		
			sTree[node<<1|1] += (e-mid)*lazy[node];		
			lazy[node] = 0;
		}
		
		int query(int i, int j)
		{
			return query(1,1,N,i,j);
		}
		
		int query(int node, int b, int e, int i, int j)	// O(log n)
		{
			if(i>e || j <b)
				return 0;		
			if(b>= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			int q1 = query(node<<1,b,mid,i,j);
			int q2 = query(node<<1|1,mid+1,e,i,j);
			return q1 + q2;	
					
		}
	
	}
	static boolean f2=false;
	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r)	//C(p x r) = A(p x q) x (q x r) -- O(p x q x r)
	{
		long[][] C = new long[p][r];
		for(int i = 0; i < p; ++i) {
			for(int j = 0; j < r; ++j) {
				for(int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j]+(a2[i][k]%mod * b[k][j]%mod))%mod;
		 C[i][j]%=mod;
				}
		
			}
		}
		return C;
	}
		
	
	
		
		
		
		

		
	
	static ArrayList<Pair> a1[];
	static int memo1[];
	
	static boolean vis[];
	static TreeSet<Integer> set=new TreeSet<Integer>();
	
	static long modPow(long ways, long count, long mod)	// O(log e)
	{
		ways %= mod;
		long res = 1;
		while(count > 0)
		{
			if((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res%mod;
	}
	
	
	

	static long gcd(long ans,long b) {
		if(b==0) {
			return ans;
		}
		return gcd(b,ans%b);
	}
	static int[] isComposite;
	static int[] valid;
	
	static ArrayList<Integer> primes;
	static ArrayList<Integer> l;
	static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
		l=new ArrayList<>();
		valid=new int[N+1];
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				l.add(i);
				valid[i]=1;
				
				for (int j = i*2; j <=N; j +=i) {	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
						
						
						
				      
					}
			}
		
		for(int i=0 ; i<primes.size() ; i++) {
			for(int j:primes) {
				if(primes.get(i)*j>N) {
					break;
				}
				valid[primes.get(i)*j]=1;
			}
		}
	}
	
	public static long[] schuffle(long[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x=(int)(Math.random()*a2.length);
		long temp=a2[x];
		a2[x]=a2[i];
		a2[i]=temp;
		}
		return a2;
	}
		static int V;
	static long INF=(long) 1E18;
		
		static class Edge2 
		{
			int node;
		long cost;
			long next;
			Edge2(int a, int c,Long long1) { node = a;	cost = long1; next=c;}
			
		}
static class Scanner {
		StringTokenizer st;
		BufferedReader br;
 
		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}
 
		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}
 
		public String nextLine() throws IOException {
			return br.readLine();
		}
 
		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}
 
		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}
 
		public char nextChar() throws IOException {
			return next().charAt(0);
		}
 
		public Long nextLong() throws IOException {
			return Long.parseLong(next());
		}
 
		public boolean ready() throws IOException {
			return br.ready();
		}
 
		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}	
}