Problem: You have a malfunctioning microwave in which you want to put some bananas. You have n time-steps before the microwave stops working completely. At each time-step, it displays a new operation.

Let k be the number of bananas in the microwave currently. Initially, k = 0. In the i-th operation, you are given three parameters t_i, x_i, y_i in the input. Based on the value of t_i, you must do one of the following:

Type 1: (t_i=1, x_i, y_i) — pick an a_i, such that 0 ≤ a_i ≤ y_i, and perform the following update a_i times: k:=⌈ (k + x_i) ⌉.

Type 2: (t_i=2, x_i, y_i) — pick an a_i, such that 0 ≤ a_i ≤ y_i, and perform the following update a_i times: k:=⌈ (k ⋅ x_i) ⌉.

Note that x_i can be a fractional value. See input format for more details. Also, ⌈ x ⌉ is the smallest integer ≥ x.

At the i-th time-step, you must apply the i-th operation exactly once.

For each j such that 1 ≤ j ≤ m, output the earliest time-step at which you can create exactly j bananas. If you cannot create exactly j bananas, output -1.

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 200) and m (2 ≤ m ≤ 10^5).

Then, n lines follow, where the i-th line denotes the operation for the i-th timestep. Each such line contains three space-separated integers t_i, x'_i and y_i (1 ≤ t_i ≤ 2, 1≤ y_i≤ m).

Note that you are given x'_i, which is 10^5 ⋅ x_i. Thus, to obtain x_i, use the formula x_i= \dfrac{x'_i} {10^5}.

For type 1 operations, 1 ≤ x'_i ≤ 10^5 ⋅ m, and for type 2 operations, 10^5 < x'_i ≤ 10^5 ⋅ m.

Output

Print m integers, where the i-th integer is the earliest time-step when you can obtain exactly i bananas (or -1 if it is impossible).

Examples

Input


3 20
1 300000 2
2 400000 2
1 1000000 3


Output


-1 -1 1 -1 -1 1 -1 -1 -1 3 -1 2 3 -1 -1 3 -1 -1 -1 3 


Input


3 20
1 399999 2
2 412345 2
1 1000001 3


Output


-1 -1 -1 1 -1 -1 -1 1 -1 -1 3 -1 -1 -1 3 -1 2 -1 3 -1 

Note

In the first sample input, let us see how to create 16 number of bananas in three timesteps. Initially, k=0.

  * In timestep 1, we choose a_1=2, so we apply the type 1 update — k := ⌈(k+3)⌉ — two times. Hence, k is now 6. 
  * In timestep 2, we choose a_2=0, hence value of k remains unchanged. 
  * In timestep 3, we choose a_3=1, so we are applying the type 1 update k:= ⌈(k+10)⌉ once. Hence, k is now 16. 



It can be shown that k=16 cannot be reached in fewer than three timesteps with the given operations.

In the second sample input, let us see how to create 17 number of bananas in two timesteps. Initially, k=0.

  * In timestep 1, we choose a_1=1, so we apply the type 1 update — k := ⌈(k+3.99999)⌉ — once. Hence, k is now 4. 
  * In timestep 2, we choose a_2=1, so we apply the type 2 update — k := ⌈(k⋅ 4.12345)⌉ — once. Hence, k is now 17. 



It can be shown that k=17 cannot be reached in fewer than two timesteps with the given operations.

Solution (in JAVA):

Solution:

import java.io.*;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.util.*;
import java.util.concurrent.LinkedBlockingDeque;
import java.io.FileWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.LinkedList;


// graph, dfs,bfs, get connected components,iscycle, isbipartite, dfs on trees



public class scratch_25{
    static class Graph{

        public static class Vertex{
            HashMap<Integer,Integer> nb= new HashMap<>();     // for neighbours of each vertex
        }

        public static HashMap<Integer,Vertex> vt;   // for vertices(all)

        public Graph(){
            vt= new HashMap<>();
        }

        public static int numVer(){
            return vt.size();
        }

        public static boolean contVer(int ver){
            return vt.containsKey(ver);
        }

        public static void addVer(int ver){
            Vertex v= new Vertex();
            vt.put(ver,v);
        }






        public static void addEdge(int ver1, int ver2, int weight){
            if(!vt.containsKey(ver1) || !vt.containsKey(ver2)){
                return;
            }
            Vertex v1= vt.get(ver1);
            Vertex v2= vt.get(ver2);
            v1.nb.put(ver2,weight);                                 // if previously there is an edge, then this replaces that edge
            v2.nb.put(ver1,weight);
        }

        public static void delEdge(int ver1, int ver2){

            if(!vt.containsKey(ver1) || !vt.containsKey(ver2)){
                return;
            }
            vt.get(ver1).nb.remove(ver2);
            vt.get(ver2).nb.remove(ver1);
        }
        public static void delVer(int ver){
            if(!vt.containsKey(ver)){
                return;
            }
            Vertex v1= vt.get(ver);
            ArrayList<Integer> arr= new ArrayList<>(v1.nb.keySet());
            for (int i = 0; i <arr.size() ; i++) {
                int s= arr.get(i);
                vt.get(s).nb.remove(ver);
            }
            vt.remove(ver);
        }
        static boolean done[];
        static int parent[];
        static ArrayList<Integer>vals= new ArrayList<>();


        public static boolean isCycle(int i){
            Stack<Integer>stk= new Stack<>();

            stk.push(i);
            while(!stk.isEmpty()){
                int x= stk.pop();
                vals.add(x);
                //   System.out.print("current="+x+" stackinit="+stk);
                if(!done[x]){
                    done[x]=true;
                }
                else if(done[x] ){
                    return true;
                }

                ArrayList<Integer>ar= new ArrayList<>(vt.get(x).nb.keySet());
                for (int j = 0; j <ar.size() ; j++) {
                    if(parent[x]!=ar.get(j)){
                        parent[ar.get(j)]=x;
                        stk.push(ar.get(j));
                    }
                }
                // System.out.println(" stackfin="+stk);
            }
            return false;
        }
        static int[]level;
        static int[] curr;
        static int[] fin;

        public static void dfs(int src){
            done[src]=true;
            level[src]=0;
            Queue<Integer>q= new LinkedList<>();
            q.add(src);
            while(!q.isEmpty()){
                int x= q.poll();
                ArrayList<Integer>arr= new ArrayList<>(vt.get(x).nb.keySet());
                for (int i = 0; i <arr.size() ; i++) {
                    int v= arr.get(i);
                    if(!done[v]){
                        level[v]=level[x]+1;
                        done[v]=true;
                        q.offer(v);
                    }
                }
            }
        }
        static int oc[];
        static int ec[];


        public static void dfs1(int src){
            Queue<Integer>q= new LinkedList<>();
            q.add(src);
            done[src]= true;
            //    int count=0;
            while(!q.isEmpty()){

                int x= q.poll();
                //   System.out.println("x="+x);
                int even= ec[x];
                int odd= oc[x];
                if(level[x]%2==0){
                    int val= (curr[x]+even)%2;
                    if(val!=fin[x]){
                        //  System.out.println("bc");
                        even++;
                        vals.add(x);
                    }
                }
                else{
                    int val= (curr[x]+odd)%2;
                    if(val!=fin[x]){
                        //   System.out.println("bc");
                        odd++;
                        vals.add(x);
                    }

                }
                ArrayList<Integer>arr= new ArrayList<>(vt.get(x).nb.keySet());
                //   System.out.println(arr);
                for (int i = 0; i <arr.size() ; i++) {
                    int v= arr.get(i);
                    if(!done[v]){
                        done[v]=true;
                        oc[v]=odd;
                        ec[v]=even;
                        q.add(v);
                    }
                }
            }

        }

        static long popu[];
        static long happy[];
        static long count[]; // total people crossing that pos
        static long sum[]; // total sum of happy people including that.


        public static void bfs(int x){
            done[x]=true;
            long total= popu[x];
//            long smile= happy[x];
            ArrayList<Integer>nbrs= new ArrayList<>(vt.get(x).nb.keySet());
            for (int i = 0; i <nbrs.size() ; i++) {
                int r= nbrs.get(i);
                if(!done[r]){
                    bfs(r);
                    total+=count[r];
//                    smile+=sum[r];
                }
            }
            count[x]=total;
//            sum[x]=smile;
        }


        public static void bfs1(int x){
            done[x]=true;
//            long total= popu[x];
            long smile= 0;
            ArrayList<Integer>nbrs= new ArrayList<>(vt.get(x).nb.keySet());
            for (int i = 0; i <nbrs.size() ; i++) {
                int r= nbrs.get(i);
                if(!done[r]){
                    bfs1(r);
//                    total+=count[r];
                    smile+=happy[r];
                }
            }
//            count[x]=total;
            sum[x]=smile;
        }


    }




    static class Reader {
        static BufferedReader reader;
        static StringTokenizer tokenizer;

        /**
         * call this method to initialize reader for InputStream
         */
        static void init(InputStream input) {
            reader = new BufferedReader(
                    new InputStreamReader(input));
            tokenizer = new StringTokenizer("");
        }

        /**
         * get next word
         */
        static String next() throws IOException {
            while (!tokenizer.hasMoreTokens()) {
                //TODO add check for eof if necessary
                tokenizer = new StringTokenizer(
                        reader.readLine());
            }
            return tokenizer.nextToken();
        }

        static int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        static double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        static long nextLong() throws IOException {
            return Long.parseLong(next());
        }
    }


    static class Pair implements Comparable<Pair>     {
        long pos;
        long val;



        public Pair(long pos, long val) {
            this.pos = pos;
            this.val = val;


        }


        @Override
        public String toString() {
            return pos + " " + val;
        }

        @Override
        public int compareTo(Pair o) {
            if(this.pos >o.pos){
                return 1;
            }
            else if(this.pos==o.pos){
                return 0;
            }
            else{
                return -1;
            }
        }


    }



    static class Node implements Comparable<Node>{
        int ind;
        int len;
        int val;
        Node prev;
        Node next;
        public Node(int ind, int len,int val){
            this.ind=ind;
            this.len=len;
            this.val=val;
        }

        @Override
        public String toString() {
            if(this.next!=null){
            return ind+ " " + len+" "+val+" "+this.next.val;
        }
            return ind+" "+len+" "+val+" "+-1;


        }

        @Override
        public int compareTo(Node o) {
            if(this.len <o.len){
                return 1;
            }
            if(this.ind==o.ind && this.len==o.len && this.val==o.val){
                return 0;
            }
             if(this.len==o.len){
                if(this.ind>o.ind){
                    return 1;
                }
                else{
                    return -1;
                }
            }
            return -1;

        }
        @Override
        public boolean equals(Object x){
            Node a= (Node)x;
            if(this.ind==a.ind && this.len==a.len && this.val==a.val){
                return true;
            }
            return false;
        }
    }


    // After writing solution, quick scan for:
    //   array out of bounds
    //   special cases e.g. n=1?
    //
    // Big numbers arithmetic bugs:
    //   int overflow
    //   sorting, or taking max, or negative after MOD


    public static void main(String[] args)  throws IOException {
        Reader.init(System.in);
        BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
            int n= Reader.nextInt();
            int m= Reader.nextInt();
            boolean done[]= new boolean[m+1];
            long min[]= new long[m+1];
            Arrays.fill(min,-1);
            done[0]=true;
            for (int i = 0; i <n ; i++) {
                int op= Reader.nextInt();
                ArrayList<Integer>memer= new ArrayList<>();
                if(op==1){
                    long x= Reader.nextLong();
                    long y= Reader.nextLong();

                    for (int j = 0; j <=m ; j++) {
                        if(done[j]){
                            long v= j;
                            for (int k = 0; k <y ; k++) {
                                v= v+ceil_div(x,100000);
                                if(v>m || done[(int)v]){
                                    break;
                                }
//                                done[(int)v]=true;
                                memer.add((int)v);
                                min[(int)v]=i+1;
                            }
                        }
                    }
                }
                else if(op==2){
                    long x= Reader.nextLong();
                    long y= Reader.nextLong();

                    for (int j = 0; j <=m ; j++) {
                        if(done[j]){
                            long v= j;
                            for (int k = 0; k <y ; k++) {
                                v= ceil_div(x*v,100000);
                                if(v>m || done[(int)v]){
                                    break;
                                }
//                                done[(int)v]=true;
                                memer.add((int)v);
                                min[(int)v]=i+1;
                            }
                        }
                    }

                }
                for (int j = 0; j <memer.size() ; j++) {
                    done[memer.get(j)]=true;
                }
            }
            
            for (int i = 1; i <m+1 ; i++) {
                out.append(min[i]+" ");
            }













        out.flush();
        out.close();
    }



    public static int localMinUtil(int[] arr, int low,
                                   int high, int n)
    {


        // Find index of middle element
        int mid = low + (high - low) / 2;

        // Compare middle element with its neighbours
        // (if neighbours exist)
        if(mid == 0 || arr[mid - 1] > arr[mid] && mid == n - 1 ||
                arr[mid] < arr[mid + 1])
            return mid;

            // If middle element is not minima and its left
            // neighbour is smaller than it, then left half
            // must have a local minima.
        else if(mid > 0 && arr[mid - 1] < arr[mid])
            return localMinUtil(arr, low, mid - 1, n);

        // If middle element is not minima and its right
        // neighbour is smaller than it, then right half
        // must have a local minima.
        return localMinUtil(arr, mid + 1, high, n);
    }

    // A wrapper over recursive function localMinUtil()
    public static int localMin(int[] arr, int n)
    {
        return localMinUtil(arr, 0, n - 1, n);
    }








    public static String convert(String s,int n){
        if(s.length()==n){
            return s;
        }
        else{
            int x= s.length();
            int v=n-x;
            String str="";
            StringBuilder s2= new StringBuilder();
            for (int i = 0; i <v ; i++) {
                s2.append('0');
            }
            StringBuilder s1= new StringBuilder(s);
            s2.append(s1);

//            str+=s;
            String q= s2.toString();
            return q;

        }
    }
    public static int lower(int arr[],int key){
        int low = 0;
        int high = arr.length-1;
        while(low < high){
            int mid = low + (high - low)/2;
            if(arr[mid] >= key){
                high = mid;
            }
            else{
                low = mid+1;
            }
        }
        return low;
    }

    public static int upper(int arr[],int key){
        int low = 0;
        int high = arr.length-1;
        while(low < high){
            int mid = low + (high - low+1)/2;
            if(arr[mid] <= key){
                low = mid;
            }
            else{
                high = mid-1;
            }
        }
        return low;
    }





    static long modExp(long a, long b, long mod) {
        //System.out.println("a is " + a + " and b is " + b);
        if (a==1) return 1;
        long ans = 1;
        while (b!=0) {
            if (b%2==1) {
                ans = (ans*a)%mod;
            }
            a = (a*a)%mod;
            b/=2;
        }
        return ans;
    }


    public static long modmul(long a, long b, long mod) {
        return b == 0 ? 0 : ((modmul(a, b >> 1, mod) << 1) % mod + a * (b & 1)) % mod;
    }

    static long sum(long n){
        //   System.out.println("lol="+ (n*(n-1))/2);
        return (n*(n+1))/2;
    }






    public static ArrayList<Integer> Sieve(int n) {
        boolean arr[]= new boolean [n+1];
        Arrays.fill(arr,true);
        arr[0]=false;
        arr[1]=false;
        for (int i = 2; i*i <=n ; i++) {
            if(arr[i]){
                for (int j = 2; j <=n/i ; j++) {
                    int u= i*j;
                    arr[u]=false;
                }}
        }
        ArrayList<Integer> ans= new ArrayList<>();
        for (int i = 0; i <n+1 ; i++) {
            if(arr[i]){
                ans.add(i);
            }
        }
        return ans;
    }




    static long power( long x, long y, long p)
    {

        long res = 1;


        x = x % p;

        if (x == 0) return 0;

        while (y > 0)
        {

            if((y & 1)==1)
                res = (res * x) % p;


            y = y >> 1;
            x = (x * x) % p;
        }
        return res;
    }

    public static long ceil_div(long a, long b){
        return (a+b-1)/b;
    }

    static long gcd(long a, long b)
    {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    static long lcm(long a, long b)
    {
        return (a*b)/gcd(a, b);
    }

}