#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class T, class U> using Pa = pair<T, U>;
template <class T> using vec = vector<T>;
template <class T> using vvec = vector<vec<T>>;

constexpr ll mod = 1e9+7;
struct mint {
    ll x;
    mint(ll x=0):x((x%mod+mod)%mod){}
    
    friend ostream &operator<<(ostream& os,const mint& a){
        return os << a.x;
    }

    friend istream &operator>>(istream& is,mint& a){
        ll t;
        is >> t;
        a = mint(t);
        return (is);
    }

    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) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(ll t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    // for prime mod
    mint inv() const {
        return pow(mod-2);
    }
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint a) const {
        mint res(*this);
        return res/=a;
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int N;
    cin >> N;
    vvec<int> Q(2);
    for(int i=0;i<N;i++){
        int t,x;
        cin >> t >> x;
        t = 1-t;
        x--;
        Q[t].push_back(x);
    }
    int n = Q[0].size(),m = Q[1].size();
    vvec<mint> dp(n+1,vec<mint>(m+1,0));
    dp[0][0] = 1;
    for(int i=0;i<N;i++){
        int cnt0 = 0,cnt1 = 0;
        for(auto& x:Q[0]) cnt0 += i>=x;
        for(auto& x:Q[1]) cnt1 += i<=x;
        for(int j=0;j<=n;j++){
            int k = i-j;
            if(k<0 || k>m) continue;
            if(j+1<=n) dp[j+1][k] += dp[j][k]*(cnt0-j);
            if(k+1<=m) dp[j][k+1] += dp[j][k]*(cnt1-(m-k-1));
        }
    }
    mint ans = 0;
    for(int j=0;j<=n;j++) for(int k=0;k<=m;k++) if(j+k==N) ans += dp[j][k];
    cout << ans << "\n";
}