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.*;
import java.util.*;
import java.lang.*;

//        .--------------.
//        | Try First One|
//        '--------------'
//                |     .--------------.
//                |     |              |
//                V     V              |
//              .--------------.       |
//              |      AC.     |<---.  |
//              '--------------'    |  |
//              (True)|  |(False)   |  |
//           .--------'  |          |  |
//           |           V          |  |
//           |  .--------------.    |  |
//           |  |   Try Again  |----'  |
//           |  '--------------'       |
//           |                         |
//           |  .--------------.       |
//           '->| Try Next One |-------'
//              '--------------'
public class Fire {
    //Start Stub
    static long startTime = System.nanoTime();

    //Globals Go Here

    //Globals End
    public static void main(String[] args) throws IOException {
        FastScanner sc = new FastScanner("Fire.in", true);
        FastWriter pw = new FastWriter("Fire.out", true);
        int N= sc.ni();
        Edge[] arr= new Edge[N];
        int dp[][]= new int[N+1][2001];
        for (int i = 0; i < N; i++) {
            int a= sc.ni(); int b= sc.ni(); int c=sc.ni();
            arr[i]= new Edge(a,b,c, i);
        }
        Arrays.sort(arr);
        for (int i = 0; i <= N; i++) {
            for (int j = 0; j < 2001; j++) {
                dp[i][j]= (int) -1e9;
            }
        }
        dp[0][0]=0;
        for (int i = 0; i < N; i++) {
            for (int j = 0; j+arr[i].ti< arr[i].di; j++) {
                if (dp[i][j] != (int) -1e9){
                    dp[i+1][j+ arr[i].ti]= Math.max(dp[i][j] + arr[i].val, dp[i+1][j+ arr[i].ti]);
                }
            }
            for (int j = 0; j < 2001; j++) {
                dp[i+1][j]= Math.max(dp[i+1][j], dp[i][j]);
            }
        }
        int best=0;
        for (int i = 0; i < 2001; i++) {
            if (dp[N][i] > dp[N][best]){
                best= i;
            }
        }
        int res= dp[N][best];
        // find the order
        Stack<Integer> list= new Stack<>();
        for (int i = N; i > 0; i--) {
            if(dp[i][best] != dp[i-1][best]) {
                list.push(arr[i-1].i+1);
                best -= arr[i-1].ti;
            }
        }
        pw.println(res);
        pw.println(list.size());
        while (list.size()>0){
            pw.print(list.peek()+ " "); list.pop();
        }
        /* End Stub */
        long endTime = System.nanoTime();
        //System.out.println("Execution Time: " + (endTime - startTime) / 1e9 + " s");
        pw.close();
    }


    static class Edge implements Comparable<Edge>{
        int ti, di, val, i;
        public Edge(int a, int b, int c, int i){
            this.ti=a;
            this.di=b;
            this.val=c;
            this.i=i;
        }
        public int compareTo(Edge obj){
            return di-obj.di;
        }
    }



    static class SegmentTree {
        public long[] arr;
        public long[] tree;
        public int N;

        //Zero initialization
        public SegmentTree(int n) {
            N = n;
            arr = new long[N];
            tree = new long[4 * N + 1];
        }

        public long query(int treeIndex, int lo, int hi, int i, int j) {
            // query for arr[i..j]
            if (lo > j || hi < i)
                return 0;
            if (i <= lo && j >= hi)
                return tree[treeIndex];
            int mid = lo + (hi - lo) / 2;

            if (i > mid)
                return query(2 * treeIndex + 2, mid + 1, hi, i, j);
            else if (j <= mid)
                return query(2 * treeIndex + 1, lo, mid, i, j);

            long leftQuery = query(2 * treeIndex + 1, lo, mid, i, mid);
            long rightQuery = query(2 * treeIndex + 2, mid + 1, hi, mid + 1, j);

            // merge query results
            return merge(leftQuery, rightQuery);
        }

        public void update(int treeIndex, int lo, int hi, int arrIndex, long val) {
            if (lo == hi) {
                tree[treeIndex] = val;
                arr[arrIndex] = val;
                return;
            }

            int mid = lo + (hi - lo) / 2;

            if (arrIndex > mid)
                update(2 * treeIndex + 2, mid + 1, hi, arrIndex, val);
            else if (arrIndex <= mid)
                update(2 * treeIndex + 1, lo, mid, arrIndex, val);

            // merge updates
            tree[treeIndex] = merge(tree[2 * treeIndex + 1], tree[2 * treeIndex + 2]);
        }

        public long merge(long a, long b) {
            return (a + b);
        }
    }

    static class djset {
        int N;
        int[] parent;

        // Creates a disjoint set of size n (0, 1, ..., n-1)
        public djset(int n) {
            parent = new int[n];
            N = n;
            for (int i = 0; i < n; i++)
                parent[i] = i;
        }

        public int find(int v) {

            // I am the club president!!! (root of the tree)
            if (parent[v] == v) return v;

            // Find my parent's root.
            int res = find(parent[v]);

            // Attach me directly to the root of my tree.
            parent[v] = res;
            return res;
        }

        public boolean union(int v1, int v2) {

            // Find respective roots.
            int rootv1 = find(v1);
            int rootv2 = find(v2);

            // No union done, v1, v2 already together.
            if (rootv1 == rootv2) return false;

            // Attach tree of v2 to tree of v1.
            parent[rootv2] = rootv1;
            return true;
        }


    }

    static class FastScanner {
        public BufferedReader br;
        public StringTokenizer st;
        public InputStream is;

        public FastScanner(String name, boolean debug) throws IOException {
            if (debug) {
                is = System.in;
            } else {
                is = new FileInputStream(name);
            }
            br = new BufferedReader(new InputStreamReader(is), 32768);
            st = null;
        }


        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }

        public int ni() {
            return Integer.parseInt(next());
        }

        public long nl() {
            return Long.parseLong(next());
        }

        public double nd() {
            return Double.parseDouble(next());
        }

        public String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
            return str;
        }
    }

    static class FastWriter {
        public PrintWriter pw;

        public FastWriter(String name, boolean debug) throws IOException {
            if (debug) {
                pw = new PrintWriter(System.out);
            } else {
                pw = new PrintWriter(new BufferedWriter(new FileWriter(new File(name))));
            }
        }

        public void println(Object text) {
            pw.println(text);
        }

        public void print(Object text) {
            pw.print(text);
        }

        public void close() {
            pw.close();
        }

        public void flush() {
            pw.flush();
        }
    }

}