#include #define rep(i,n) for(int i=0;i<(int)(n);i++) #define FOR(i,n,m) for(int i=(int)(n); i<=(int)(m); i++) #define RFOR(i,n,m) for(int i=(int)(n); i>=(int)(m); i--) #define ITR(x,c) for(__typeof(c.begin()) x=c.begin();x!=c.end();x++) #define RITR(x,c) for(__typeof(c.rbegin()) x=c.rbegin();x!=c.rend();x++) #define setp(n) fixed << setprecision(n) template bool chmax(T &a, const T &b) { if (a bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; } #define ll long long #define vll vector #define vi vector #define pll pair #define pi pair #define all(a) (a.begin()),(a.end()) #define rall(a) (a.rbegin()),(a.rend()) #define fi first #define se second #define pb push_back #define ins insert #define debug(a) cerr<<(a)< ostream &operator<<(ostream &os, const pair &p){return os<<"("< istream &operator>>(istream &is, pair &p){return is>>p.fi>>p.se;} //------------------------------------------------- //--Strongly Connected Components //------------------------------------------------- class SCC { private: using SCCGraph = vector >; int N,sz; SCCGraph G, rG; stack st; vector seen; vector belong; void dfs(int v){ seen[v] = true; for(auto u:G[v]){ if (seen[u]) continue; dfs(u); } st.push(v); } void rdfs(int v, int k){ seen[v] = true; belong[v] = k; for(auto u:rG[v]){ if (seen[u]) continue; rdfs(u,k); } } public: SCC(int n):N(n),G(n),rG(n),seen(n),belong(n){} void add_edge(int u, int v){ G[u].push_back(v); rG[v].push_back(u); } void build(){ for(int v=0; v > get_list(){ vector > ret(sz); for(int v=0; v solve(){ scc.build(); vector ret(N); for(int i=0; i v) ret[i] = true; else if(u < v) ret[i] = false; else return vector(); } return ret; } }; struct Range {int l,r;}; bool cross(Range a, Range b){ if (a.l>b.l) swap(a,b); return (b.l<=a.r); } //------------------------------------------------- int main(void) { cin.tie(0); ios::sync_with_stdio(false); int N,M; cin>>N>>M; auto rev = [&](const Range &rg){ return Range{M-rg.r-1, M-rg.l-1}; }; vector rs(N); rep(i,N){ int l,r; cin>>l>>r; rs[i] = Range{l,r}; } TwoSAT ts(N); rep(i,N)FOR(j,i+1,N-1){ rep(f,2)rep(g,2){ Range a = f?rev(rs[i]):rs[i]; Range b = g?rev(rs[j]):rs[j]; if (cross(a,b)) ts.add_clause(i,f,j,g); } } auto ans = ts.solve(); cout<<(ans.empty()?"NO":"YES")<<"\n"; return 0; }