Problem: You are given n arrays that can have different sizes. You also have a table with w columns and n rows. The i-th array is placed horizontally in the i-th row. You can slide each array within its row as long as it occupies several consecutive cells and lies completely inside the table.

You need to find the maximum sum of the integers in the j-th column for each j from 1 to w independently.

<image> Optimal placements for columns 1, 2 and 3 are shown on the pictures from left to right.

Note that you can exclude any array out of a column provided it remains in the window. In this case its value is considered to be zero.

Input

The first line contains two integers n (1 ≤ n ≤ 10^{6}) and w (1 ≤ w ≤ 10^{6}) — the number of arrays and the width of the table.

Each of the next n lines consists of an integer l_{i} (1 ≤ l_{i} ≤ w), the length of the i-th array, followed by l_{i} integers a_{i1}, a_{i2}, …, a_{il_i} (-10^{9} ≤ a_{ij} ≤ 10^{9}) — the elements of the array.

The total length of the arrays does no exceed 10^{6}.

Output

Print w integers, the i-th of them should be the maximum sum for column i.

Examples

Input


3 3
3 2 4 8
2 2 5
2 6 3


Output


10 15 16 


Input


2 2
2 7 8
1 -8


Output


7 8 

Note

Illustration for the first example is in the statement.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

public class E {
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int n = scanner.nextInt();
        int w = scanner.nextInt();
        long[] pref = new long[w+1];
        long[] ans = new long[w+1];
        for(int i = 0; i < n; i++) {
            long mx =0;
            int l = scanner.nextInt();
            long[] arr = new long[l+1];
            for(int j = 1; j <=l; j++) {
                arr[j] = scanner.nextInt();
                mx = Math.max(mx, arr[j]);
            }
            //can be accessed regardless of location in the array
            if (w > 2 * l) {
                pref[l+1] += mx;
                pref[w-l+1] -= mx;
            }
            //Maintaining a max deque
            ArrayDeque<Pair> queue = new ArrayDeque<>();
            for(int j = 1; j <=l; j++) {
                //no longer in the active range
                while(!queue.isEmpty() && queue.peekFirst().a <= j) queue.pollFirst();
                //not optimal
                while(!queue.isEmpty() && queue.peekLast().b <= arr[j]) queue.pollLast();
                //add the elt
                queue.offerLast(new Pair(w-l+j+1, arr[j]));
                //if negative and can choose to ignore the elt, do so
                if (queue.peekFirst().b < 0 && j + l <= w) ans[j] += 0;
                //ow just update normally
                else ans[j] += queue.peekFirst().b;
            }
            //run the same ordeal backwards
            queue = new ArrayDeque<>();
            //need to care to aviod overlap
            for(int j = w; j>w-l && j > l; j--) {
                while(!queue.isEmpty() && queue.peekFirst().a >= j) queue.pollFirst();
                while(!queue.isEmpty() && queue.peekLast().b <= arr[l-(w-j)]) queue.pollLast();
                queue.offerLast(new Pair(l-(w-j)-1, arr[l-(w-j)]));
                if (queue.peekFirst().b < 0 && j-l>0) ans[j]+=0;
                else ans[j] += queue.peekFirst().b;
            }
        }
        for(int i = 1; i <=w; i++) {
            pref[i] += pref[i-1];
            ans[i] += pref[i];
            if (i > 1) out.print(" ");
            out.print(ans[i]);
        }
        out.println();
        out.flush();
    }
    static class Pair {
        int a; long b;
        public Pair(int aa, long bb) {
            a = aa; b = bb;
        }
    }
    
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;
        
        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }
        
        public FastScanner() {
            this(new InputStreamReader(System.in));
        }
        
        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        
        int nextInt() {
            return Integer.parseInt(next());
        }
        
        long nextLong() {
            return Long.parseLong(next());
        }
        
        double nextDouble() {
            return Double.parseDouble(next());
        }
        
        String readNextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}