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 Rust (1.42.0) (judge verdict: Time Limit Exceeded)

pub trait IterScan: Sized {
    type Output;
    fn scan<'a, I: Iterator<Item = &'a str>>(iter: &mut I) -> Option<Self::Output>;
}
pub trait MarkedIterScan: Sized {
    type Output;
    fn mscan<'a, I: Iterator<Item = &'a str>>(self, iter: &mut I) -> Option<Self::Output>;
}
#[derive(Debug)]
pub struct Scanner<'a> {
    iter: std::str::SplitAsciiWhitespace<'a>,
}
impl<'a> Scanner<'a> {
    #[inline]
    pub fn new(s: &'a str) -> Self {
        let iter = s.split_ascii_whitespace();
        Self { iter }
    }
    #[inline]
    pub fn scan<T: IterScan>(&mut self) -> <T as IterScan>::Output {
        T::scan(&mut self.iter).unwrap()
    }
    #[inline]
    pub fn mscan<T: MarkedIterScan>(&mut self, marker: T) -> <T as MarkedIterScan>::Output {
        marker.mscan(&mut self.iter).unwrap()
    }
    #[inline]
    pub fn scan_vec<T: IterScan>(&mut self, size: usize) -> Vec<<T as IterScan>::Output> {
        (0..size)
            .map(|_| T::scan(&mut self.iter).unwrap())
            .collect()
    }
    #[inline]
    pub fn scan_chars(&mut self) -> Vec<char> {
        self.iter.next().unwrap().chars().collect::<Vec<char>>()
    }
}
mod scanner_impls {
    use super::*;
    macro_rules !iter_scan_impls {($($t :ty ) *) =>{$(impl IterScan for $t {type Output =Self ;#[inline ] fn scan <'a ,I :Iterator <Item =&'a str >>(iter :&mut I ) ->Option <Self >{iter .next () ?.parse ::<$t >() .ok () } } ) *} ;}
    iter_scan_impls !(char u8 u16 u32 u64 usize i8 i16 i32 i64 isize f32 f64 u128 i128 String ) ;
    macro_rules !iter_scan_tuple_impl {($($T :ident ) *) =>{impl <$($T :IterScan ) ,*>IterScan for ($($T ,) *) {type Output =($(<$T as IterScan >::Output ,) *) ;#[inline ] fn scan <'a ,It :Iterator <Item =&'a str >>(_iter :&mut It ) ->Option <Self ::Output >{Some (($($T ::scan (_iter ) ?,) *) ) } } } ;}
    iter_scan_tuple_impl!();
    iter_scan_tuple_impl!(A);
    iter_scan_tuple_impl !(A B ) ;
    iter_scan_tuple_impl !(A B C ) ;
    iter_scan_tuple_impl !(A B C D ) ;
    iter_scan_tuple_impl !(A B C D E ) ;
    iter_scan_tuple_impl !(A B C D E F ) ;
    iter_scan_tuple_impl !(A B C D E F G ) ;
    iter_scan_tuple_impl !(A B C D E F G H ) ;
    iter_scan_tuple_impl !(A B C D E F G H I ) ;
    iter_scan_tuple_impl !(A B C D E F G H I J ) ;
    iter_scan_tuple_impl !(A B C D E F G H I J K ) ;
    pub struct ScannerIter<'a, 'b, T> {
        inner: &'b mut Scanner<'a>,
        _marker: std::marker::PhantomData<fn() -> T>,
    }
    impl<'a, 'b, T: IterScan> Iterator for ScannerIter<'a, 'b, T> {
        type Item = <T as IterScan>::Output;
        fn next(&mut self) -> Option<Self::Item> {
            T::scan(&mut self.inner.iter)
        }
    }
    impl<'a> Scanner<'a> {
        #[inline]
        pub fn iter<'b, T: IterScan>(&'b mut self) -> ScannerIter<'a, 'b, T> {
            ScannerIter {
                inner: self,
                _marker: std::marker::PhantomData,
            }
        }
    }
}
mod marker {
    use super::*;
    pub struct Usize1;
    impl IterScan for Usize1 {
        type Output = usize;
        #[inline]
        fn scan<'a, I: Iterator<Item = &'a str>>(iter: &mut I) -> Option<Self::Output> {
            usize::scan(iter).map(|x| x.wrapping_sub(1))
        }
    }
    pub struct Isize1;
    impl IterScan for Isize1 {
        type Output = isize;
        #[inline]
        fn scan<'a, I: Iterator<Item = &'a str>>(iter: &mut I) -> Option<Self::Output> {
            isize::scan(iter).map(|x| x.wrapping_sub(1))
        }
    }
}
#[macro_export]
macro_rules !min {($e :expr ) =>{$e } ;($e :expr ,$($es :expr ) ,+) =>{std ::cmp ::min ($e ,min !($($es ) ,+) ) } ;}
#[macro_export]
macro_rules !chmin {($dst :expr ,$($src :expr ) ,+) =>{{let x =std ::cmp ::min ($dst ,min !($($src ) ,+) ) ;$dst =x ;} } ;}
#[macro_export]
macro_rules !max {($e :expr ) =>{$e } ;($e :expr ,$($es :expr ) ,+) =>{std ::cmp ::max ($e ,max !($($es ) ,+) ) } ;}
#[macro_export]
macro_rules !chmax {($dst :expr ,$($src :expr ) ,+) =>{{let x =std ::cmp ::max ($dst ,max !($($src ) ,+) ) ;$dst =x ;} } ;}
fn main() {
    #[allow(unused_imports)]
    use std::io::{Read as _, Write as _};
    let __out = std::io::stdout();
    let mut __in_buf = String::new();
    std::io::stdin().read_to_string(&mut __in_buf).unwrap();
    let mut __scanner = Scanner::new(&__in_buf);
    #[allow(unused_macros)]
    macro_rules !scan {() =>{scan !(usize ) } ;(($($t :tt ) ,*) ) =>{($(scan !($t ) ) ,*) } ;([$t :tt ;$len :expr ] ) =>{(0 ..$len ) .map (|_ |scan !($t ) ) .collect ::<Vec <_ >>() } ;({$t :tt =>$f :expr } ) =>{$f (scan !($t ) ) } ;(chars ) =>{__scanner .scan_chars () } ;($t :ty ) =>{__scanner .scan ::<$t >() } ;}
    let mut __out = std::io::BufWriter::new(__out.lock());
    #[allow(unused_macros)]
    macro_rules !print {($($arg :tt ) *) =>(::std ::write !(__out ,$($arg ) *) .unwrap () ) }
    #[allow(unused_macros)]
    macro_rules !println {($($arg :tt ) *) =>(::std ::writeln !(__out ,$($arg ) *) .unwrap () ) }
    #[allow(unused_macros)]
    macro_rules! echo {
        ($iter :expr ) => {
            echo!($iter, "\n")
        };
        ($iter :expr ,$sep :expr ) => {
            let mut iter = $iter;
            if let Some(item) = iter.next() {
                print!("{}", item);
            }
            for item in iter {
                print!("{}{}", $sep, item);
            }
            println!();
        };
    }
    let n = scan!();
    let xyp = scan!([(i64, i64, i64); n]);
    let mut xv = vec![vec![std::i64::MAX; n]; 1 << n];
    let mut yv = vec![vec![std::i64::MAX; n]; 1 << n];
    let mut ans = vec![std::i64::MAX; n + 1];
    for i in 0..1 << n {
        for j in 0..n {
            let (x, y, p) = xyp[j];
            xv[i][j] = x.abs() * p;
            yv[i][j] = y.abs() * p;
            for k in 0..n {
                let (xx, yy, _) = xyp[k];
                if i & 1 << k != 0 {
                    chmin!(xv[i][j], (x - xx).abs() * p);
                    chmin!(yv[i][j], (y - yy).abs() * p);
                }
            }
        }
    }
    for i in 0..1 << n {
        let mut c = 0;
        for j in 0..n {
            c += (i & 1 << j != 0) as usize;
        }
        for j in (0..=i).rev() {
            let j = j & i;
            let mut a = 0;
            for k in 0..n {
                if i & 1 << k == 0 {
                    a += min!(xv[j][k], yv[i - j][k]);
                }
            }
            chmin!(ans[c], a);
        }
    }
    echo!(ans.into_iter());
}