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)

macro_rules ! input { ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut iter = $ s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; ( $ ( $ r : tt ) * ) => { let s = { use std :: io :: Read ; let mut s = String :: new ( ) ; std :: io :: stdin ( ) . read_to_string ( & mut s ) . unwrap ( ) ; s } ; let mut iter = s . split_whitespace ( ) ; input_inner ! { iter , $ ( $ r ) * } } ; }
macro_rules ! input_inner { ( $ iter : expr ) => { } ; ( $ iter : expr , ) => { } ; ( $ iter : expr , mut $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let mut $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , mut $ var : ident $ ( $ r : tt ) * ) => { let mut $ var = $ iter . next ( ) . unwrap ( ) . parse ::< usize > ( ) . unwrap ( ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ iter , $ t ) ; input_inner ! { $ iter $ ( $ r ) * } } ; ( $ iter : expr , $ var : ident $ ( $ r : tt ) * ) => { let $ var = $ iter . next ( ) . unwrap ( ) . parse ::< usize > ( ) . unwrap ( ) ; input_inner ! { $ iter $ ( $ r ) * } } ; }
macro_rules ! read_value { ( $ iter : expr , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ iter , $ t ) ) ,* ) } ; ( $ iter : expr , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ iter , $ t ) ) . collect ::< Vec < _ >> ( ) } ; ( $ iter : expr , chars ) => { read_value ! ( $ iter , String ) . chars ( ) . collect ::< Vec < char >> ( ) } ; ( $ iter : expr , usize1 ) => { read_value ! ( $ iter , usize ) - 1 } ; ( $ iter : expr , $ t : ty ) => { $ iter . next ( ) . unwrap ( ) . parse ::<$ t > ( ) . unwrap ( ) } ; }
const MOD: i64 = 1_000_000_007;
#[derive(Copy, Clone, Debug, PartialEq, Eq, PartialOrd, Ord)]
struct Modi64(i64);
impl Modi64 {
    #[inline]
    fn new(i: i64) -> Modi64 {
        Modi64(Modi64::modulo(i, MOD))
    }
    #[inline]
    fn modulo(i: i64, m: i64) -> i64 {
        if i < 0 {
            i % m + m
        } else {
            i % m
        }
    }
    fn pow(self, y: i64) -> Modi64 {
        let mut y = Modi64(Modi64::modulo(y, MOD - 1));
        let mut x = 1.into();
        let mut base = self;
        while y.0 > 0 {
            if y.0 & 1 == 1 {
                x *= base;
            }
            base *= base;
            y = Modi64(y.0 >> 1);
        }
        x
    }
    #[inline]
    fn inv(self) -> Modi64 {
        self.pow(MOD - 2)
    }
}
impl From<i64> for Modi64 {
    fn from(i: i64) -> Modi64 {
        Modi64::new(i)
    }
}
impl std::ops::Add for Modi64 {
    type Output = Modi64;
    fn add(self, rhs: Modi64) -> Modi64 {
        (self.0 + rhs.0).into()
    }
}
impl std::ops::AddAssign for Modi64 {
    fn add_assign(&mut self, rhs: Modi64) {
        *self = *self + rhs;
    }
}
impl std::ops::Sub for Modi64 {
    type Output = Modi64;
    fn sub(self, rhs: Modi64) -> Modi64 {
        (self.0 - rhs.0).into()
    }
}
impl std::ops::SubAssign for Modi64 {
    fn sub_assign(&mut self, rhs: Modi64) {
        *self = *self - rhs;
    }
}
impl std::ops::Mul for Modi64 {
    type Output = Modi64;
    fn mul(self, rhs: Modi64) -> Modi64 {
        (self.0 * rhs.0).into()
    }
}
impl std::ops::MulAssign for Modi64 {
    fn mul_assign(&mut self, rhs: Modi64) {
        *self = *self * rhs;
    }
}
impl std::ops::Div for Modi64 {
    type Output = Modi64;
    fn div(self, rhs: Modi64) -> Modi64 {
        self * rhs.inv()
    }
}
impl std::ops::DivAssign for Modi64 {
    fn div_assign(&mut self, rhs: Modi64) {
        *self = *self / rhs;
    }
}
impl std::ops::Neg for Modi64 {
    type Output = Modi64;
    fn neg(self) -> Modi64 {
        (-self.0).into()
    }
}
fn main() {
    input! { n, a: [[u8; n]; n] };
    let mut dp = vec![Modi64(0); 1 << n];
    dp[0] = 1.into();
    for i in 0..(1 << n) {
        for j in 0..n {
            let jb = 1 << j;
            let c = (i as usize).count_ones() as usize;
            if i & jb != 0 || a[c][j] == 0 {
                continue;
            }
            let d = dp[i];
            dp[i | jb] += d;
        }
    }
    println!("{}", dp[(1 << n) - 1].0);
}