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.util.*;
import java.io.*;
import java.math.*;
public class smack1 {
	InputStream is;
	PrintWriter out;
	static long mod=pow(10,9)+7;
	Pair p[];
	void solve()
	{
		int n=ni();
		p=new Pair[n];
		for(int i=0;i<n;i++)
			p[i]=new Pair(ni(),ni(),ni(),i);
		Arrays.sort(p);
		dp=new int[n][2000+1];
		for(int i=0;i<n;i++)
			Arrays.fill(dp[i],-1);
		path=new Pair[n][2000+1];
		out.println(rec(0,0));
		int i=0,j=0,count=0;
		while(i<n&&j<2001)
		{
		    if(path[i][j]==null)
		        break;
			if(path[i][j].k==1)
				count++;
			int temp=i;
			i=path[i][j].x;
			j=path[temp][j].y;
		}
		out.println(count);
		i=0;
		j=0;
		while(i<n&&j<2001)
		{
		    if(path[i][j]==null)
		        break;
			if(path[i][j].k==1)
				out.print((path[i][j].i+1)+" ");
			int temp=i;
			i=path[i][j].x;
			j=path[temp][j].y;
		}
	}
	int dp[][];
	Pair path[][];
	int rec(int n,int time)
	{
		if(n==p.length)
			return 0;
		if(n>p.length)
			return Integer.MIN_VALUE;
		if(dp[n][time]!=-1)
			return dp[n][time];
		dp[n][time]=0;
		if(time+p[n].x<p[n].y)
		{
			dp[n][time]=Math.max(dp[n][time],rec(n+1,time+p[n].x)+p[n].k);
			path[n][time]=new Pair(n+1,time+p[n].x,1,p[n].i);
		}
		int val=rec(n+1,time);
		if(val>dp[n][time])
		{
			dp[n][time]=Math.max(dp[n][time],val);
			path[n][time]=new Pair(n+1,time,-1,p[n].i);
		}
		return dp[n][time];
	}
	static long d, x, y;
	void extendedEuclid(long A, long B) {
	    if(B == 0) {
	        d = A;
	        x = 1;
	        y = 0;
	    }
	    else {
	        extendedEuclid(B, A%B);
	        long temp = x;
	        x = y;
	        y = temp - (A/B)*y;
	    }
	}
	long modInverse(long A,long M) //A and M are coprime
	{
	    extendedEuclid(A,M);
	    return (x%M+M)%M;    //x may be negative
	}
	public static void mergeSort(int[] arr, int l ,int r){
		if((r-l)>=1){
			int mid = (l+r)/2;
			mergeSort(arr,l,mid);
			mergeSort(arr,mid+1,r);
			merge(arr,l,r,mid);
		}
	}
	public static void merge(int arr[], int l, int r, int mid){
		int n1 = (mid-l+1), n2 = (r-mid);
		int left[] = new int[n1];
		int right[] = new int[n2];
		for(int i =0 ;i<n1;i++) left[i] = arr[l+i];
		for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i];
		int i =0, j =0, k = l;
		while(i<n1 && j<n2){
			if(left[i]>right[j]){
				arr[k++] = right[j++];
			}
			else{
				arr[k++] = left[i++];
			}
		}
		while(i<n1) arr[k++] = left[i++];
		while(j<n2) arr[k++] = right[j++];
	}
	public static void mergeSort(long[] arr, int l ,int r){
		if((r-l)>=1){
			int mid = (l+r)/2;
			mergeSort(arr,l,mid);
			mergeSort(arr,mid+1,r);
			merge(arr,l,r,mid);
		}
	}
	public static void merge(long arr[], int l, int r, int mid){
		int n1 = (mid-l+1), n2 = (r-mid);
		long left[] = new long[n1];
		long right[] = new long[n2];
		for(int i =0 ;i<n1;i++) left[i] = arr[l+i];
		for(int i =0 ;i<n2;i++) right[i] = arr[mid+1+i];
		int i =0, j =0, k = l;
		while(i<n1 && j<n2){
			if(left[i]>right[j]){
				arr[k++] = right[j++];
			}
			else{
				arr[k++] = left[i++];
			}
		}
		while(i<n1) arr[k++] = left[i++];
		while(j<n2) arr[k++] = right[j++];
	}
	 static class Pair implements Comparable<Pair>{
		 
	       int x,y,k,i;
	        
		Pair (int x,int y,int k,int i){
			this.x=x;
			this.y=y;
			this.k=k;
			this.i=i;
		}
	        
		public int compareTo(Pair o) {
			return this.y-o.y;
		}
	     public boolean equals(Object o) {
	           if (o instanceof Pair) {
	                Pair p = (Pair)o;
	                return p.x == x && p.y == y && p.k==k;
	            }
	            return false;
	        }
	           
	
	        @Override
	        public String toString() {
	            return  "("+x + " " + y +" "+k+" "+i+" )";
	        }
	    
	    } 
	    
	    public static boolean isPal(String s){
	        for(int i=0, j=s.length()-1;i<=j;i++,j--){
	                if(s.charAt(i)!=s.charAt(j)) return false;
	        }
	        return true;
	    }
	    public static String rev(String s){
			StringBuilder sb=new StringBuilder(s);
			sb.reverse();
			return sb.toString();
	    }
	    
	    public static long gcd(long x,long y){
		if(x%y==0)
			return y;
		else
			return gcd(y,x%y);
	    }
	    
	    public static int gcd(int x,int y){
		if(x%y==0)
			return y;
		else 
			return gcd(y,x%y);
	    }
	    
	    public static long gcdExtended(long a,long b,long[] x){
	        
	        if(a==0){
	            x[0]=0;
	            x[1]=1;
	            return b;
	        }
	        long[] y=new long[2];
	        long gcd=gcdExtended(b%a, a, y);
	        
	        x[0]=y[1]-(b/a)*y[0];
	        x[1]=y[0];
	        
	        return gcd;
	    }
	    
	    public static int abs(int a,int b){
		return (int)Math.abs(a-b);
	    }
	 
	    public static long abs(long a,long b){
		return (long)Math.abs(a-b);
	    }
	    
	    public static int max(int a,int b){
		if(a>b)
			return a;
		else
			return b;
	    }
	 
	    public static int min(int a,int b){
		if(a>b)
			return b;
		else 
			return a;
	    }
	    
	    public static long max(long a,long b){
		if(a>b)
			return a;
		else
			return b;
	    }
	 
	    public static long min(long a,long b){
		if(a>b)
			return b;
		else 
			return a;
	    }
	 
	    public static long pow(long n,long p,long m){
		 long  result = 1;
		  if(p==0)
		    return 1;
		if (p==1)
		    return n;
		while(p!=0)
		{
		    if(p%2==1)
		        result *= n;
		    if(result>=m)
		    result%=m;
		    p >>=1;
		    n*=n;
		    if(n>=m)
		    n%=m;
		}
		return result;
	    }
	    
	    public static long pow(long n,long p){
		long  result = 1;
		  if(p==0)
		    return 1;
		if (p==1)
		    return n;
		while(p!=0)
		{
		    if(p%2==1)
		        result *= n;	    
		    p >>=1;
		    n*=n;	    
		}
		return result;
	    }
	    public static void debug(Object... o) {
			System.out.println(Arrays.deepToString(o));
		}
	    void run() throws Exception {
			is = System.in;
			out = new PrintWriter(System.out);
			solve();
			out.flush();
		}
	   
	    public static void main(String[] args) throws Exception {
			new Thread(null, new Runnable() {
				public void run() {
					try {
						new smack1().run();
					} catch (Exception e) {
						e.printStackTrace();
					}
				}
			}, "1", 1 << 26).start();
		}
	    private byte[] inbuf = new byte[1024];
		public int lenbuf = 0, ptrbuf = 0;
	 
		private int readByte() {
			if (lenbuf == -1)
				throw new InputMismatchException();
			if (ptrbuf >= lenbuf) {
				ptrbuf = 0;
				try {
					lenbuf = is.read(inbuf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (lenbuf <= 0)
					return -1;
			}
			return inbuf[ptrbuf++];
		}
	 
		private boolean isSpaceChar(int c) {
			return !(c >= 33 && c <= 126);
		}
	 
		private int skip() {
			int b;
			while ((b = readByte()) != -1 && isSpaceChar(b));
			return b;
		}
	 
		private double nd() {
			return Double.parseDouble(ns());
		}
	 
		private char nc() {
			return (char) skip();
		}
	 
		private String ns() {
			int b = skip();
			StringBuilder sb = new StringBuilder();
			while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != ' ')
				sb.appendCodePoint(b);
				b = readByte();
			}
			return sb.toString();
		}
	 
		private char[] ns(int n) {
			char[] buf = new char[n];
			int b = skip(), p = 0;
			while (p < n && !(isSpaceChar(b))) {
				buf[p++] = (char) b;
				b = readByte();
			}
			return n == p ? buf : Arrays.copyOf(buf, p);
		}
	 
		private char[][] nm(int n, int m) {
			char[][] map = new char[n][];
			for (int i = 0; i < n; i++)
				map[i] = ns(m);
			return map;
		}
	 
		private int[] na(int n) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++)
				a[i] = ni();
			return a;
		}
	 
		private int ni() {
			int num = 0, b;
			boolean minus = false;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}
	 
			while (true) {
				if (b >= '0' && b <= '9') {
					num = num * 10 + (b - '0');
				} else {
					return minus ? -num : num;
				}
				b = readByte();
			}
		}
	 
		private long nl() {
			long num = 0;
			int b;
			boolean minus = false;
			while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
				;
			if (b == '-') {
				minus = true;
				b = readByte();
			}
	 
			while (true) {
				if (b >= '0' && b <= '9') {
					num = num * 10 + (b - '0');
				} else {
					return minus ? -num : num;
				}
				b = readByte();
			}
		}
	 
}