Problem:
Score: 500 points

Problem Statement

New AtCoder City has an infinite grid of streets, as follows:

At the center of the city stands a clock tower. Let (0, 0) be the coordinates of this point.

A straight street, which we will call East-West Main Street, runs east-west and passes the clock tower. It corresponds to the x-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to East-West Main Street, with a distance of 1 between them. They correspond to the lines \ldots, y = -2, y = -1, y = 1, y = 2, \ldots in the two-dimensional coordinate plane.

A straight street, which we will call North-South Main Street, runs north-south and passes the clock tower. It corresponds to the y-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to North-South Main Street, with a distance of 1 between them. They correspond to the lines \ldots, x = -2, x = -1, x = 1, x = 2, \ldots in the two-dimensional coordinate plane.

There are N residential areas in New AtCoder City. The i-th area is located at the intersection with the coordinates (X_i, Y_i) and has a population of P_i. Each citizen in the city lives in one of these areas.

The city currently has only two railroads, stretching infinitely, one along East-West Main Street and the other along North-South Main Street.

M-kun, the mayor, thinks that they are not enough for the commuters, so he decides to choose K streets and build a railroad stretching infinitely along each of those streets.

Let the walking distance of each citizen be the distance from his/her residential area to the nearest railroad.

M-kun wants to build railroads so that the sum of the walking distances of all citizens, S, is minimized.

For each K = 0, 1, 2, \dots, N, what is the minimum possible value of S after building railroads?

Constraints

1 \leq N \leq 15

-10 \ 000 \leq X_i \leq 10 \ 000

-10 \ 000 \leq Y_i \leq 10 \ 000

1 \leq P_i \leq 1 \ 000 \ 000

The locations of the N residential areas, (X_i, Y_i), are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 Y_1 P_1
X_2 Y_2 P_2
: : :
X_N Y_N P_N

Output

Print the answer in N+1 lines.

The i-th line (i = 1, \ldots, N+1) should contain the minimum possible value of S after building railroads for the case K = i-1.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

8
2 2 286017
3 1 262355
2 -2 213815
1 -3 224435
-2 -2 136860
-3 -1 239338
-2 2 217647
-1 3 141903

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.lang.reflect.Field;
import java.util.Arrays;
import java.util.NoSuchElementException;
import java.util.function.IntUnaryOperator;
import java.util.function.LongUnaryOperator;


public class Main {
    public static void main(String[] args) throws Exception {
        Field f = System.out.getClass().getDeclaredField("autoFlush");
        f.setAccessible(true);
        f.set(System.out, false);
        execute(System.in, System.out);
    }

    public static void execute(InputStream in, OutputStream out) {
        ExtendedScanner s = new ExtendedScanner(System.in);
        Out o = new Out(System.out);
        solve(s, o);
        o.flush();
        o.close();
    }

    public static void solve(ExtendedScanner sc, Out out) {
        int n = sc.nextInt();
        long[][] xyp = new long[n][3];
        for (int i = 0; i < n; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            int p = sc.nextInt();
            xyp[i][0] = x;
            xyp[i][1] = y;
            xyp[i][2] = p;
        }
        long[][] xs = new long[1 << n][];
        long[][] ys = new long[1 << n][];
        for (int i = 0; i < 1 << n; i++) {
            int l = Integer.bitCount(i);
            xs[i] = new long[l + 3];
            ys[i] = new long[l + 3];
            for (int j = 0, idx = 0; j < n; j++) {
                if (((i >> j) & 1) == 1) {
                    xs[i][idx] = xyp[j][0];
                    ys[i][idx] = xyp[j][1];
                    idx++;
                }
            }
            xs[i][l] = ys[i][l] = 0;
            xs[i][l + 1] = ys[i][l + 1] = Const.IINF;
            xs[i][l + 2] = ys[i][l + 2] = -Const.IINF;
            Arrays.sort(xs[i]);
            Arrays.sort(ys[i]);
        }
        long[][] sx = new long[n + 1][1 << n];
        long[][] sy = new long[n + 1][1 << n];
        for (int i = 0; i <= n; i++) {
            Arrays.fill(sx[i], Const.IINF);
            Arrays.fill(sy[i], Const.IINF);
        }
        for (int j = 0; j < 1 << n; j++) {
            for (int k = j;; k = (k - 1) & j) {
                int i = Integer.bitCount(k);
                long tx = 0, ty = 0;
                for (int l = 0; l < n; l++) {
                    if (((j >> l) & 1) == 0) continue;
                    int xl = 0, xr = xs[k].length;
                    while (xr - xl > 1) {
                        int xm = (xl + xr) >> 1;
                        if (xyp[l][0] >= xs[k][xm]) xl = xm;
                        else xr = xm;
                    }
                    tx += xyp[l][2] * Math.min(xs[k][xr] - xyp[l][0], xyp[l][0] - xs[k][xl]);
                    int yl = 0, yr = ys[k].length;
                    while (yr - yl > 1) {
                        int ym = (yl + yr) >> 1;
                        if (xyp[l][1] >= ys[k][ym]) yl = ym;
                        else yr = ym;
                    }
                    ty += xyp[l][2] * Math.min(ys[k][yr] - xyp[l][1], xyp[l][1] - ys[k][yl]);
                }
                sx[i][j] = Math.min(sx[i][j], tx);
                sy[i][j] = Math.min(sy[i][j], ty);
                if (k == 0) break;
            }
        }
        for (int k = 0; k <= n; k++) {
            long min = Const.IINF;
            for (int i = 0; i < 1 << n; i++) {
                for (int a = 0; a <= k; a++) {
                    min = Math.min(min, sx[a][i] + sy[k - a][((1 << n) - 1) ^ i]);
                }
            }
            out.write(min).write('\n');
        }
    }
}


/**
 * @author https://atcoder.jp/users/suisen
 */
class BasicScanner {
    private final InputStream in;
    private final byte[] buf = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    public BasicScanner(InputStream in) {this.in = in;}
    private boolean hasNextByte() {
        if (ptr < buflen) return true;
        ptr = 0;
        try {buflen = in.read(buf);}
        catch (final IOException e) {e.printStackTrace();}
        return buflen > 0;
    }
    private int readByte() {return hasNextByte() ? buf[ptr++] : -1;}
    public boolean hasNext() {
        while (hasNextByte() && !(33 <= buf[ptr] && buf[ptr] <= 126)) ptr++;
        return hasNextByte();
    }
    private StringBuilder nextSequence() {
        if (!hasNext()) throw new NoSuchElementException();
        final StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (33 <= b && b <= 126) {sb.appendCodePoint(b); b = readByte();}
        return sb;
    }
    public char nextChar() {
        return (char) readByte();
    }
    public String next() {
        return nextSequence().toString();
    }
    public String next(int len) {
        return new String(nextChars(len));
    }
    public char[] nextChars() {
        final StringBuilder sb = nextSequence();
        int l = sb.length();
        char[] dst = new char[l];
        sb.getChars(0, l, dst, 0);
        return dst;
    }
    public char[] nextChars(int len) {
        if (!hasNext()) throw new NoSuchElementException();
        char[] s = new char[len];
        int i = 0;
        int b = readByte();
        while (33 <= b && b <= 126 && i < len) {s[i++] = (char) b; b = readByte();}
        if (i != len) throw new NoSuchElementException(String.format("length %d is longer than the sequence.", len));
        return s;
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {minus = true; b = readByte();}
        if (b < '0' || '9' < b) throw new NumberFormatException();
        while (true) {
            if ('0' <= b && b <= '9') n = n * 10 + b - '0';
            else if (b == -1 || !(33 <= b && b <= 126)) return minus ? -n : n;
            else throw new NumberFormatException();
            b = readByte();
        }
    }
    public int nextInt() {return Math.toIntExact(nextLong());}
    public double nextDouble() {return Double.parseDouble(next());}
}


/**
 * @author https://atcoder.jp/users/suisen
 */
class Out {
    private final OutputStream out;
    private byte[] buf = new byte[1024];
    private int ptr = 0;
    private static final int AUTO_FLUSH_THRETHOLD = 1 << 17;

    public Out(OutputStream out) {
        this.out = out;
    }

    public void flush() {
        try {
            out.write(buf, 0, ptr);
            out.flush();
            ptr = 0;
        } catch (IOException e) {e.printStackTrace();}
    }

    public void close() {
        try {out.close();} catch (IOException e) {e.printStackTrace();}
    }

    public Out write(Object o) {
        return write(o.toString());
    }

    public Out write(String s) {
        try {
            Field strValueField = s.getClass().getDeclaredField("value");
            strValueField.setAccessible(true);
            byte[] b = (byte[]) strValueField.get(s);
            int l = s.length();
            if (l > AUTO_FLUSH_THRETHOLD) {
                flush();
                int i = 0;
                while (i + AUTO_FLUSH_THRETHOLD < l) {
                    out.write(b, i, AUTO_FLUSH_THRETHOLD);
                    out.flush();
                    i += AUTO_FLUSH_THRETHOLD;
                }
                ensureCapacity(l - i);
                System.arraycopy(b, i, buf, 0, l - i);
                ptr = l - i;
            } else {
                ensureCapacity(ptr + l);
                System.arraycopy(b, 0, buf, ptr, l);
                ptr += l;
            }
        } catch (Exception e) {e.printStackTrace();}
        return this;
    }

    public Out write(char[] c) {
        int l = c.length;
        if (l > AUTO_FLUSH_THRETHOLD) {
            flush();
            ensureCapacity(AUTO_FLUSH_THRETHOLD);
            int i = 0;
            while (i + AUTO_FLUSH_THRETHOLD < l) {
                for (int j = 0; j < AUTO_FLUSH_THRETHOLD; j++) {
                    buf[ptr++] = (byte) c[i + j];
                }
                flush();
                i += AUTO_FLUSH_THRETHOLD;
            }
            for (; i < l; i++) {
                buf[ptr++] = (byte) c[i];
            }
        } else {
            ensureCapacity(ptr + l);
            for (char ch : c) buf[ptr++] = (byte) ch;
        }
        return this;
    }

    public Out write(char c) {
        ensureCapacity(ptr + 1);
        buf[ptr++] = (byte) c;
        return this;
    }

    public Out write(byte b) {
        ensureCapacity(ptr + 1);
        buf[ptr++] = b;
        return this;
    }

    public Out write(int x) {
        if (x == 0) {
            ensureCapacity(ptr + 1);
            buf[ptr++] = '0';
            return this;
        }
        int d = stringSize(x);
        ensureCapacity(ptr + d);
        if (x < 0) {
            buf[ptr++] = '-';
            x = -x;
            d--;
        }
        int j = ptr + d; 
        while (x > 0) {
            buf[--j] = (byte) ('0' + (x % 10));
            x /= 10;
        }
        ptr += d;
        return this;
    }

    public Out write(long x) {
        if (x == 0) {
            ensureCapacity(ptr + 1);
            buf[ptr++] = '0';
            return this;
        }
        int d = stringSize(x);
        ensureCapacity(ptr + d);
        if (x < 0) {
            buf[ptr++] = '-';
            x = -x;
            d--;
        }
        int j = ptr + d; 
        while (x > 0) {
            buf[--j] = (byte) ('0' + (x % 10));
            x /= 10;
        }
        ptr += d;
        return this;
    }

    public Out write(double d) {
        return write(Double.toString(d));
    }

    private void ensureCapacity(int cap) {
        if (cap > AUTO_FLUSH_THRETHOLD) {
            flush();
        }
        if (cap >= buf.length) {
            int newLength = buf.length;
            while (newLength < cap) newLength <<= 1;
            byte[] newBuf = new byte[newLength];
            System.arraycopy(buf, 0, newBuf, 0, buf.length);
            buf = newBuf;
        }
    }
    private static int stringSize(long x) {
        int d = 1;
        if (x >= 0) {d = 0; x = -x;}
        long p = -10;
        for (int i = 1; i < 19; i++, p *= 10) if (x > p) return i + d;
        return 19 + d;
    }
    private static int stringSize(int x) {
        int d = 1;
        if (x >= 0) {d = 0; x = -x;}
        int p = -10;
        for (int i = 1; i < 10; i++, p *= 10) if (x > p) return i + d;
        return 10 + d;
    }
}


/**
 * @author https://atcoder.jp/users/suisen
 */
final class ExtendedScanner extends BasicScanner {
    public ExtendedScanner(InputStream in) {super(in);}
    public int[] ints(final int n) {
        final int[] a = new int[n];
        Arrays.setAll(a, $ -> nextInt());
        return a;
    }
    public int[] ints(final int n, final IntUnaryOperator f) {
        final int[] a = new int[n];
        Arrays.setAll(a, $ -> f.applyAsInt(nextInt()));
        return a;
    }
    public int[][] ints(final int n, final int m) {
        final int[][] a = new int[n][];
        Arrays.setAll(a, $ -> ints(m));
        return a;
    }
    public int[][] ints(final int n, final int m, final IntUnaryOperator f) {
        final int[][] a = new int[n][];
        Arrays.setAll(a, $ -> ints(m, f));
        return a;
    }
    public long[] longs(final int n) {
        final long[] a = new long[n];
        Arrays.setAll(a, $ -> nextLong());
        return a;
    }
    public long[] longs(final int n, final LongUnaryOperator f) {
        final long[] a = new long[n];
        Arrays.setAll(a, $ -> f.applyAsLong(nextLong()));
        return a;
    }
    public long[][] longs(final int n, final int m) {
        final long[][] a = new long[n][];
        Arrays.setAll(a, $ -> longs(m));
        return a;
    }
    public long[][] longs(final int n, final int m, final LongUnaryOperator f) {
        final long[][] a = new long[n][];
        Arrays.setAll(a, $ -> longs(m, f));
        return a;
    }
    public char[][] charArrays(final int n) {
        final char[][] c = new char[n][];
        Arrays.setAll(c, $ -> nextChars());
        return c;
    }
    public char[][] charArrays(final int n, final int m) {
        final char[][] c = new char[n][];
        Arrays.setAll(c, $ -> nextChars(m));
        return c;
    }
    public double[] doubles(final int n) {
        final double[] a = new double[n];
        Arrays.setAll(a, $ -> nextDouble());
        return a;
    }
    public double[][] doubles(final int n, final int m) {
        final double[][] a = new double[n][];
        Arrays.setAll(a, $ -> doubles(m));
        return a;
    }
    public String[] strings(final int n) {
        final String[] s = new String[n];
        Arrays.setAll(s, $ -> next());
        return s;
    }
    public String[] strings(final int n, final int m) {
        final String[] s = new String[n];
        Arrays.setAll(s, $ -> next(m));
        return s;
    }
    public boolean[][] grid(final int h, final int w, final char trueCharacter) {
        final boolean[][] grid = new boolean[h][w];
        for (int i = 0; i < h; i++) {
            final char[] s = nextChars(w);
            for (int j = 0; j < w; j++) grid[i][j] = s[j] == trueCharacter;
        }
        return grid;
    }
}

/**
 * @author https://atcoder.jp/users/suisen
 */
class Const {
    public static final long   LINF = 1l << 59;
    public static final int    IINF = 1  << 30;
    public static final double DINF = 1e150;

    public static final double SDELTA = 1e-12;
    public static final double DELTA  = 1e-9;
    public static final double LDELTA = 1e-6;

    public static final int[] dx8 = {1, 0, -1, 0, 1, -1, -1, 1};
    public static final int[] dy8 = {0, 1, 0, -1, 1, 1, -1, -1};
    public static final int[] dx4 = {1, 0, -1, 0};
    public static final int[] dy4 = {0, 1, 0, -1};

    private Const(){}
}