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: Runtime Error)

#include <bits/stdc++.h>
#define fi first
#define se second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define drep(i,n) for(int i = (n)-1; i >= 0; --i)
#define srep(i,s,t) for (int i = s; i < t; ++i)
#define rng(a) a.begin(),a.end()
#define maxs(x,y) (x = max(x,y))
#define mins(x,y) (x = min(x,y))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x = max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define sz(x) (int)(x).size()
#define pcnt __builtin_popcountll
#define uni(x) x.erase(unique(rng(x)),x.end())
#define snuke srand((unsigned)clock()+(unsigned)time(NULL));
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T> >
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) v(v(T))
using namespace std;
typedef long long int ll;
typedef unsigned uint;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<P> vp;
inline int in() { int x; scanf("%d",&x); return x;}
template<typename T>inline istream& operator>>(istream&i,v(T)&v)
{rep(j,sz(v))i>>v[j];return i;}
template<typename T>string join(const v(T)&v)
{stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template<typename T>inline ostream& operator<<(ostream&o,const v(T)&v)
{if(sz(v))o<<join(v);return o;}
template<typename T1,typename T2>inline istream& operator>>(istream&i,pair<T1,T2>&v)
{return i>>v.fi>>v.se;}
template<typename T1,typename T2>inline ostream& operator<<(ostream&o,const pair<T1,T2>&v)
{return o<<v.fi<<","<<v.se;}
template<typename T>inline ll suma(const v(T)& a) { ll res(0); for (auto&& x : a) res += x; return res;}
const double eps = 1e-10;
const ll LINF = 1001002003004005006ll;
const int INF = 1001001001;
#define dame { puts("-1"); return 0;}
#define yn {puts("YES");}else{puts("NO");}
const int MX = 305;

// Mod int
const int mod = 1000000007;
struct mint {
  ll x;
  mint():x(0){}
  mint(ll x):x((x%mod+mod)%mod){}
  // mint(ll x):x(x){}
  mint& fix() { x = (x%mod+mod)%mod; return *this;}
  mint operator-() const { return mint(0) - *this;}
  mint& operator+=(const mint& a){ if((x+=a.x)>=mod) x-=mod; return *this;}
  mint& operator-=(const mint& a){ if((x+=mod-a.x)>=mod) x-=mod; return *this;}
  mint& operator*=(const mint& a){ (x*=a.x)%=mod; return *this;}
  mint& operator/=(const mint& a){ (x*=a.ex(mod-2).x)%=mod; return *this;}
  mint operator+(const mint& a)const{ return mint(*this) += a;}
  mint operator-(const mint& a)const{ return mint(*this) -= a;}
  mint operator*(const mint& a)const{ return mint(*this) *= a;}
  mint operator/(const mint& a)const{ return mint(*this) /= a;}
  mint ex(ll t) const {
    if(!t) return 1;
    mint res = ex(t/2);
    res *= res;
    return (t&1)?res*x:res;
  }
  bool operator<(const mint& a)const{ return x < a.x;}
  bool operator==(const mint& a)const{ return x == a.x;}
};
mint ex(mint x, ll t) { return x.ex(t);}
istream& operator>>(istream&i,mint&a){i>>a.x;return i;}
ostream& operator<<(ostream&o,const mint&a){o<<a.x;return o;}
typedef vector<mint> vm;
struct comb {
  vm f, g;
  comb(){}
  comb(int mx):f(mx+1),g(mx+1) {
    f[0] = 1;
    rrep(i,mx) f[i] = f[i-1]*i;
    g[mx] = f[mx].ex(mod-2);
    for(int i=mx;i>0;i--) g[i-1] = g[i]*i;
  }
  mint c(int a, int b) {
    if (a < b) return 0;
    return f[a]*g[b]*g[a-b];
  }
};
//

mint dp[MX][MX][MX];

// mint d[MX][MX];
// mint f(int n) {
//   d[0][0] = 1;
//   rep(i,n) {
//     int ni = i+1;
//     rep(j,ni) {
//       d[i+1][j+1] += d[i][j];
//       if (j) d[i+1][j-1] += d[i][j];
//     }
//   }
//   mint res = d[n][0];
//   rrep(i,n/2) res *= i;
//   return res;
// }

int main() {
  int n;
  scanf("%d",&n);
  int m = n<<1;
  vi ab(m);
  cin>>ab;
  map<int,int> mp;
  rrep(i,m) mp[i] = 0;
  vi x;
  rep(i,n) {
    int a = ab[i*2];
    int b = ab[i*2+1];
    if (a > b) swap(a,b);
    if (b == -1) continue;
    if (a == -1) {
      x.pb(b);
      continue;
    }
    mp.erase(a);
    mp.erase(b);
  }

  m = 0;
  for (auto&& p : mp) {
    p.se = m++;
  }

  rep(i,sz(x)) x[i] = mp[x[i]];

  n = m/2;
  vi s(m);
  rep(i,sz(x)) s[x[i]] = 1;

  dp[0][0][0] = 1;
  rep(i,m) {
    int ni = i+1;

    if (!s[i]) {
      rep(j,ni)rep(k,ni) {
        mint now = dp[i][j][k];
        dp[i+1][j][k+1] += now;
        if (j) dp[i+1][j-1][k] += now;
      }
      rep(j,ni)rep(k,ni) {
        mint now = dp[i][j][k];
        dp[i+1][j+1][k] += now;
        // if (k) dp[i+1][j][k-1] += now*k;
      }
      continue;
    }

    rep(j,ni)rep(k,ni) {
      mint now = dp[i][j][k];
      dp[i+1][j+1][k] += now;
      if (k) dp[i+1][j][k-1] += now*k;
    }
  }

  int r = (m-sz(x)*2)/2;
  mint ans = dp[m][0][0];
  rrep(i,r) ans *= i;
  cout<<ans<<endl;
  return 0;
}