#include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define REP(i,n) for(ll (i)=0;(i)<(n);(i)++) #define rep(i,j,n) for(ll (i)=(j);(i)<(n);(i)++) #define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i) #define ll long long #define ull unsigned long long #define all(hoge) (hoge).begin(),(hoge).end() #define en '\n' typedef pair P; const long long INF = 1LL << 60; const int INF_INT = 1 << 25; const long long MOD = 1e9 + 7; typedef vector Array; typedef vector Matrix; template using vec = vector; template using vvec = vector>; const int loose = 0; const int tight = 1; template inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; } template inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; } //グラフ関連 struct Edge {//グラフ ll to, cap, rev; Edge(ll _to, ll _cap, ll _rev) { to = _to; cap = _cap; rev = _rev; } }; typedef vector Edges; typedef vector Graph; void add_edge(Graph& G, ll from, ll to, ll cap, bool revFlag, ll revCap) { G[from].push_back(Edge(to, cap, (ll)G[to].size())); if (revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1)); } int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); ll n; cin >> n; vec

xt(n); ll con = 0; REP(i, n) { ll t,x; cin >> t >> x; if (t == 1) { //x以上の条件のときは以下の場合になおす x--; con++; } xt[i] = { x,t }; } sort(all(xt));//小さい方からだと、使える個数が確定する Matrix dp(n+1, Array(n+1, 0));//i番目までで、j個の制約を除外した場合 dp[0][0] = 1; REP(i, n) { ll t = xt[i].second; ll x = xt[i].first; REP(j, i + 1) { ll num = max(x - i + j, 0LL);//使える個数 if (t == 0) { (dp[i + 1][j] += dp[i][j] * num % MOD) %= MOD; } else { (dp[i + 1][j] += dp[i][j] * num % MOD) %= MOD; (dp[i + 1][j + 1] += dp[i][j]) %= MOD;//i個目の制約条件を除外 } } } //除外した制約の分の余った数字は自由に並べられる->i! ll fact = 1; REP(i, n+1) { (dp[n][i] *= fact) %= MOD; (fact *= i + 1) %= MOD; } //包除原理 ll ans = 0; REP(i, con+1) { if (i % 2 == 0) { (ans += dp[n][con - i]) %= MOD;//制約を偶数個含む場合は足す } else { (ans += MOD - dp[n][con - i]) %= MOD;//制約を奇数個含む場合は引く } } cout << ans << en; return 0; }