Problem:
Score : 100 points

Problem Statement

There are N men and N women, both numbered 1, 2, \ldots, N.

For each i, j (1 \leq i, j \leq N), the compatibility of Man i and Woman j is given as an integer a_{i, j}.
If a_{i, j} = 1, Man i and Woman j are compatible; if a_{i, j} = 0, they are not.

Taro is trying to make N pairs, each consisting of a man and a woman who are compatible.
Here, each man and each woman must belong to exactly one pair.

Find the number of ways in which Taro can make N pairs, modulo 10^9 + 7.

Constraints

All values in input are integers.

1 \leq N \leq 21

a_{i, j} is 0 or 1.

Input

Input is given from Standard Input in the following format:

N
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}

Output

Print the number of ways in which Taro can make N pairs, modulo 10^9 + 7.

Sample Input 1

3
0 1 1
1 0 1
1 1 1

Sample Output 1

3

There are three ways to make pairs, as follows ((i, j) denotes a pair of Man i and Woman j):

(1, 2), (2, 1), (3, 3)

(1, 2), (2, 3), (3, 1)

(1, 3), (2, 1), (3, 2)

Sample Input 2

4
0 1 0 0
0 0 0 1
1 0 0 0
0 0 1 0

Sample Output 2

1

There is one way to make pairs, as follows:

(1, 2), (2, 4), (3, 1), (4, 3)

Sample Input 3

1
0

Sample Output 3

0

Sample Input 4

21
0 0 0 0 0 0 0 1 1 0 1 1 1 1 0 0 0 1 0 0 1
1 1 1 0 0 1 0 0 0 1 0 0 0 0 1 1 1 0 1 1 0
0 0 1 1 1 1 0 1 1 0 0 1 0 0 1 1 0 0 0 1 1
0 1 1 0 1 1 0 1 0 1 0 0 1 0 0 0 0 0 1 1 0
1 1 0 0 1 0 1 0 0 1 1 1 1 0 0 0 0 0 0 0 0
0 1 1 0 1 1 1 0 1 1 1 0 0 0 1 1 1 1 0 0 1
0 1 0 0 0 1 0 1 0 0 0 1 1 1 0 0 1 1 0 1 0
0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 1 1 1 1
0 0 1 0 0 1 0 0 1 0 1 1 0 0 1 0 1 0 1 1 1
0 0 0 0 1 1 0 0 1 1 1 0 0 0 0 1 1 0 0 0 1
0 1 1 0 1 1 0 0 1 1 0 0 0 1 1 1 1 0 1 1 0
0 0 1 0 0 1 1 1 1 0 1 1 0 1 1 1 0 0 0 0 1
0 1 1 0 0 1 1 1 1 0 0 0 1 0 1 1 0 1 0 1 1
1 1 1 1 1 0 0 0 0 1 0 0 1 1 0 1 1 1 0 0 1
0 0 0 1 1 0 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1
1 0 1 1 0 1 0 1 0 0 1 0 0 1 1 0 1 0 1 1 0
0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1 1 0 0 1
0 0 0 1 0 0 1 1 0 1 0 1 0 1 1 0 0 1 1 0 1
0 0 0 0 1 1 1 0 1 0 1 1 1 0 1 1 0 0 1 1 0
1 1 0 1 1 0 0 1 1 0 1 1 0 1 1 1 1 1 0 1 0
1 0 0 1 1 0 1 1 1 1 1 0 1 0 1 1 0 0 0 0 0

Sample Output 4

102515160

Be sure to print the number modulo 10^9 + 7.

Solution in Rust (1.15.1) (judge verdict: Accepted)

#[doc = " https://github.com/hatoo/competitive-rust-snippets"]
#[allow(unused_imports)]
use std::cmp::{max, min, Ordering};
#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
#[allow(unused_imports)]
use std::io::{stdin, stdout, BufWriter, Write};
#[allow(unused_imports)]
use std::iter::FromIterator;
#[macro_export]
macro_rules ! dvec { ( $ t : expr ; $ len : expr ) => { vec ! [ $ t ; $ len ] } ; ( $ t : expr ; $ len : expr , $ ( $ rest : expr ) ,* ) => { vec ! [ dvec ! ( $ t ; $ ( $ rest ) ,* ) ; $ len ] } ; }
#[allow(unused_macros)]
macro_rules ! debug { ( $ ( $ a : expr ) ,* ) => { eprintln ! ( concat ! ( $ ( stringify ! ( $ a ) , " = {:?}, " ) ,* ) , $ ( $ a ) ,* ) ; } }
#[macro_export]
macro_rules ! input { ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut parser = Parser :: from_str ( $ s ) ; input_inner ! { parser , $ ( $ r ) * } } ; ( parser = $ parser : ident , $ ( $ r : tt ) * ) => { input_inner ! { $ parser , $ ( $ r ) * } } ; ( new_stdin_parser = $ parser : ident , $ ( $ r : tt ) * ) => { let stdin = std :: io :: stdin ( ) ; let reader = std :: io :: BufReader :: new ( stdin . lock ( ) ) ; let mut $ parser = Parser :: new ( reader ) ; input_inner ! { $ parser , $ ( $ r ) * } } ; ( $ ( $ r : tt ) * ) => { input ! { new_stdin_parser = parser , $ ( $ r ) * } } ; }
#[macro_export]
macro_rules ! input_inner { ( $ parser : ident ) => { } ; ( $ parser : ident , ) => { } ; ( $ parser : ident , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ parser , $ t ) ; input_inner ! { $ parser $ ( $ r ) * } } ; }
#[macro_export]
macro_rules ! read_value { ( $ parser : ident , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ parser , $ t ) ) ,* ) } ; ( $ parser : ident , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ parser , $ t ) ) . collect ::< Vec < _ >> ( ) } ; ( $ parser : ident , chars ) => { read_value ! ( $ parser , String ) . chars ( ) . collect ::< Vec < char >> ( ) } ; ( $ parser : ident , usize1 ) => { read_value ! ( $ parser , usize ) - 1 } ; ( $ parser : ident , $ t : ty ) => { $ parser . next ::<$ t > ( ) . expect ( "Parse error" ) } ; }
use std::io;
use std::io::BufRead;
use std::str;
pub struct Parser<R> {
    reader: R,
    buf: Vec<u8>,
    pos: usize,
}
impl Parser<io::Empty> {
    pub fn from_str(s: &str) -> Parser<io::Empty> {
        Parser {
            reader: io::empty(),
            buf: s.as_bytes().to_vec(),
            pos: 0,
        }
    }
}
impl<R: BufRead> Parser<R> {
    pub fn new(reader: R) -> Parser<R> {
        Parser {
            reader: reader,
            buf: vec![],
            pos: 0,
        }
    }
    pub fn update_buf(&mut self) {
        self.buf.clear();
        self.pos = 0;
        loop {
            let (len, complete) = {
                let buf2 = self.reader.fill_buf().unwrap();
                self.buf.extend_from_slice(buf2);
                let len = buf2.len();
                if len == 0 {
                    break;
                }
                (len, buf2[len - 1] <= 0x20)
            };
            self.reader.consume(len);
            if complete {
                break;
            }
        }
    }
    pub fn next<T: str::FromStr>(&mut self) -> Result<T, T::Err> {
        loop {
            let mut begin = self.pos;
            while begin < self.buf.len() && (self.buf[begin] <= 0x20) {
                begin += 1;
            }
            let mut end = begin;
            while end < self.buf.len() && (self.buf[end] > 0x20) {
                end += 1;
            }
            if begin != self.buf.len() {
                self.pos = end;
                return str::from_utf8(&self.buf[begin..end]).unwrap().parse::<T>();
            } else {
                self.update_buf();
            }
        }
    }
}
#[allow(unused_macros)]
macro_rules ! debug { ( $ ( $ a : expr ) ,* ) => { eprintln ! ( concat ! ( $ ( stringify ! ( $ a ) , " = {:?}, " ) ,* ) , $ ( $ a ) ,* ) ; } }
#[doc = " https://github.com/hatoo/competitive-rust-snippets"]
const BIG_STACK_SIZE: bool = true;
#[allow(dead_code)]
fn main() {
    use std::thread;
    if BIG_STACK_SIZE {
        thread::Builder::new()
            .stack_size(32 * 1024 * 1024)
            .name("solve".into())
            .spawn(solve)
            .unwrap()
            .join()
            .unwrap();
    } else {
        solve();
    }
}
const mo: i64 = 1_000_000_007;
struct Rec {
    N: usize,
    a: Vec<Vec<i64>>,
    dp: Vec<Vec<Option<i64>>>
}
impl Rec {
    fn f(&mut self, i: usize, used: u64) -> i64 {
        if i == self.N { return 1 }

        if let Some(x) = self.dp[i][used as usize] {
            return x
        }

        let mut sum = 0;
        for j in 0..self.N {
            let mask = 1<<j;
            if self.a[i][j]==1 && used & mask == 0 {
                sum += self.f(i+1, used | mask);
                sum %= mo;
            }
        }

        self.dp[i][used as usize] = Some(sum);
        sum 
    }
}
fn solve() {
    input!{
        N:usize,
        a:[[i64;N];N],
    }
    let mut dp = dvec![None; N, 1<<N];
    let mut rec = Rec {
        N: N,
        a: a,
        dp: dp
    };
    println!("{}", rec.f(0,0));
}