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:
//Implemented By Aman Kotiyal Date:-30-Mar-2021  Time:-1:57:09 pm
import java.io.*;
import java.util.*;

public class ques1
{
	public static void main(String[] args)throws Exception{ new ques1().run();} 
	long mod=1000000000+7;
	long maxn=100000;
	void solve() throws Exception
	{
		int n=ni();
		int m=ni();
		long a[][]=new long[n][3];
		for(int i=0;i<n;i++)
		{
				a[i][0]=nl();
				a[i][1]=nl();
				a[i][2]=nl();
		}
		long dp[]=new long[m+1];
		Arrays.fill(dp, 500);
		dp[0]=0;
		
		for(int i=0;i<n;i++)
		{
			HashMap<Integer,Integer> map= new HashMap<>();
			for(int j=0;j<=m;j++)
			{
				if(dp[j]>=500)continue;
				long x=j;
				for(long k=1;k<=a[i][2];k++)
				{
					if(a[i][0]==1)
					{
						x=((x*maxn + a[i][1])-1l+maxn)/maxn;
						if(x>m||dp[(int)x]!=500)break;
						map.put((int)x, i+1);
					}
					else
					{
						x=((x*a[i][1])-1l+maxn)/maxn;
						if(x>m||dp[(int)x]!=500)break;
						map.put((int)x, i+1);
					}
				}
			}
			for(Integer key:map.keySet())
				dp[key]=map.get(key);
		}
		
		for(int i=1;i<=m;i++)
		{
			if(dp[i]>=500)
				out.print("-1 ");
			else
			out.print(dp[i]+" ");
		}
		out.println();
	}
	
	/*FAST INPUT OUTPUT & METHODS BELOW*/
	
	private byte[] buf=new byte[1024];
	private int index;
	private InputStream in;
	private int total;
	private SpaceCharFilter filter;
	PrintWriter out;
	
	int min(int... ar){int min=Integer.MAX_VALUE;for(int i:ar)min=Math.min(min, i);return min;}
	long min(long... ar){long min=Long.MAX_VALUE;for(long i:ar)min=Math.min(min, i);return min;}
	int max(int... ar) {int max=Integer.MIN_VALUE;for(int i:ar)max=Math.max(max, i);return max;}
	long max(long... ar) {long max=Long.MIN_VALUE;for(long i:ar)max=Math.max(max, i);return max;}
	void reverse(int a[]){for(int i=0;i<a.length>>1;i++){int tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
	void reverse(long a[]){for(int i=0;i<a.length>>1;i++){long tem=a[i];a[i]=a[a.length-1-i];a[a.length-1-i]=tem;}}
	String reverse(String s){StringBuilder sb=new StringBuilder(s);sb.reverse();return sb.toString();}
	
	void shuffle(int a[]) {
		ArrayList<Integer> al = new ArrayList<>();
		for(int i=0;i<a.length;i++) 
			al.add(a[i]);
		
		Collections.sort(al);
		for(int i=0;i<a.length;i++) 
			a[i]=al.get(i);
	}
	long lcm(long a,long b)
	{
		return (a*b)/(gcd(a,b));
	}
	
	int gcd(int a, int b) 
	{ 
		if (a == 0) 
			return b; 
		return gcd(b%a, a); 
	} 
	long gcd(long a, long b) 
	{ 
		if (a == 0) 
			return b; 
		return gcd(b%a, a); 
	}
	/* for (1/a)%mod = ( a^(mod-2) )%mod  ----> use expo to calc -->(a^(mod-2)) */
	long expo(long p,long q)  /*  (p^q)%mod   */
	{
		long z = 1;
		while (q>0) {
			if (q%2 == 1) {
				z = (z * p)%mod;
			}
			p = (p*p)%mod;
			q >>= 1;
		}
		return z;
	}
	void run()throws Exception
	{
		in=System.in; out = new PrintWriter(System.out);
		solve();
		out.flush();
	}
	private int scan()throws IOException
	{
		if(total<0)
			throw new InputMismatchException();
		if(index>=total)
		{
			index=0;
			total=in.read(buf);
			if(total<=0)
				return -1;
		}
		return buf[index++];
	}
	private int ni() throws IOException 
	{
		int c = scan();
		while (isSpaceChar(c))
			c = scan();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = scan();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = scan();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
	private long nl() throws IOException 
	{
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = scan()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = scan();
		}
		
		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = scan();
		}
	}
	private double nd() throws IOException{
		return Double.parseDouble(ns());
	}
	private String ns() throws IOException {
		int c = scan();
		while (isSpaceChar(c))
			c = scan();
		StringBuilder res = new StringBuilder();
		do {
			if (Character.isValidCodePoint(c))
				res.appendCodePoint(c);
			c = scan();
		} while (!isSpaceChar(c));
		return res.toString();
	}
	private String nss() throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		return br.readLine();
	}
	private char nc() throws IOException 
	{
		int c = scan();
		while (isSpaceChar(c))
			c = scan();
		return (char) c;
	}
	private boolean isWhiteSpace(int n)
	{
		if(n==' '||n=='\n'||n=='\r'||n=='\t'||n==-1)
			return true;
		return false;
	}
	private boolean isSpaceChar(int c) {
		if (filter != null)
			return filter.isSpaceChar(c);
		return isWhiteSpace(c);
	}
	private interface SpaceCharFilter {
		public boolean isSpaceChar(int ch);
	}
}