Problem:
Score : 1600 points

Problem Statement

There is a sequence of length 2N: A_1, A_2, ..., A_{2N}.
Each A_i is either -1 or an integer between 1 and 2N (inclusive). Any integer other than -1 appears at most once in {A_i}.

For each i such that A_i = -1, Snuke replaces A_i with an integer between 1 and 2N (inclusive), so that {A_i} will be a permutation of 1, 2, ..., 2N.
Then, he finds a sequence of length N, B_1, B_2, ..., B_N, as B_i = min(A_{2i-1}, A_{2i}).

Find the number of different sequences that B_1, B_2, ..., B_N can be, modulo 10^9 + 7.

Constraints

1 \leq N \leq 300

A_i = -1 or 1 \leq A_i \leq 2N.

If A_i \neq -1, A_j \neq -1, then A_i \neq A_j. (i \neq j)

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_{2N}

Output

Print the number of different sequences that B_1, B_2, ..., B_N can be, modulo 10^9 + 7.

Sample Input 1

3
1 -1 -1 3 6 -1

Sample Output 1

5

There are six ways to make {A_i} a permutation of 1, 2, ..., 2N; for each of them, {B_i} would be as follows:

(A_1, A_2, A_3, A_4, A_5, A_6) = (1, 2, 4, 3, 6, 5): (B_1, B_2, B_3) = (1, 3, 5)

(A_1, A_2, A_3, A_4, A_5, A_6) = (1, 2, 5, 3, 6, 4): (B_1, B_2, B_3) = (1, 3, 4)

(A_1, A_2, A_3, A_4, A_5, A_6) = (1, 4, 2, 3, 6, 5): (B_1, B_2, B_3) = (1, 2, 5)

(A_1, A_2, A_3, A_4, A_5, A_6) = (1, 4, 5, 3, 6, 2): (B_1, B_2, B_3) = (1, 3, 2)

(A_1, A_2, A_3, A_4, A_5, A_6) = (1, 5, 2, 3, 6, 4): (B_1, B_2, B_3) = (1, 2, 4)

(A_1, A_2, A_3, A_4, A_5, A_6) = (1, 5, 4, 3, 6, 2): (B_1, B_2, B_3) = (1, 3, 2)

Thus, there are five different sequences that B_1, B_2, B_3 can be.

Sample Input 2

4
7 1 8 3 5 2 6 4

Sample Output 2

1

Sample Input 3

10
7 -1 -1 -1 -1 -1 -1 6 14 12 13 -1 15 -1 -1 -1 -1 20 -1 -1

Sample Output 3

9540576

Sample Input 4

20
-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 6 -1 -1 -1 -1 -1 7 -1 -1 -1 -1 -1 -1 -1 -1 -1 34 -1 -1 -1 -1 31 -1 -1 -1 -1 -1 -1 -1 -1

Sample Output 4

374984201

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
void ade(int &x,int y){x+=y;if(x>=mod)x-=mod;}
int n;
int nu[660];
int m;
bool book[660];
bool isx[660];
int a,b;
int dp[660][330][330];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n*2;++i)scanf("%d",&nu[i]);
	m=n*2;
	for(int i=1;i<=n;++i)
		if(nu[i*2]!=-1&&nu[i*2-1]!=-1)
		{
			m-=2;
			book[nu[i*2]]=book[nu[i*2-1]]=1;
		}
		else if(nu[i*2]==-1&&nu[i*2-1]==-1)a++;
		else
		{
			b++;
			isx[nu[i*2]+nu[i*2-1]+1]=1;
		}
	dp[0][0][0]=1;
	int x;
	int p=0;
	int v;
	for(int i=0;i<m;++i)
	{
		while(book[n*2-p])p++;
		x=n*2-p;p++;
		for(int j=0;j<=i&&j<=a+b;++j)
			for(int k=0;k<=i&&k<=b;++k)
			{
				if(!dp[i][j][k])continue;
				v=dp[i][j][k];
				if(isx[x])
				{
					if(j) ade(dp[i+1][j-1][k],v);
					ade(dp[i+1][j][k+1],v);
				}
				else
				{
					if(j) ade(dp[i+1][j-1][k],v);
					ade(dp[i+1][j+1][k],v);
					if(k) ade(dp[i+1][j][k-1],1ll*k*v%mod);
				}
			}
	}
	int fac=1;
	for(int i=1;i<=a;++i)fac=1ll*fac*i%mod;
	printf("%lld\n",1ll*fac*dp[m][0][0]%mod);
	return 0;
}