#include #define rep(i, a, n) for(int i = a; i < (n); i++) using namespace std; using ll = long long; using P = pair; const int INF = 1001001001; const ll LINF = 1001002003004005006ll; //const int mod = 1000000007; const int mod = 998244353; //MINT struct mint { unsigned x; mint(): x(0) {} mint(ll x):x((x%mod+mod)%mod) {} mint operator-() const { return mint(0) - *this;} mint operator~() const { return mint(1) / *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=(unsigned long long)x*a.x%mod; return *this;} mint& operator/=(const mint& a) { x=(unsigned long long)x*a.pow(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 pow(ll t) const { if (!t) return 1; mint res = pow(t>>1); 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;} bool operator!=(const mint& a) const { return x != a.x;} }; mint ex(mint x, ll t) { return x.pow(t);} istream& operator>>(istream& i, mint& a) { unsigned long long t; i>>t; a=mint(t); return i;} ostream& operator<<(ostream& o, const mint& a) { return o< struct BIT{ int _n; vector data; BIT(): _n(0) {}; explicit BIT(int n): _n(n), data(n) {}; void add(int p, T x){ assert(0 <= p && p < _n); p++; while(p <= _n){ data[p-1] += x; p += p&-p; } } T sum(int r){ T s = 0; while(r > 0){ s += data[r-1]; r -= r&-r; } return s; } T sum(int l, int r){ assert(0 <= l && l <= r && r <= _n); return sum(r) - sum(l); } }; // POWER_MODver. N^k % MOD ll mod_pow(ll n, ll k){ ll res = 1; for(; k > 0; k >>= 1){ if(k&1) res = (res*n)%mod; n = (n*n)%mod; } return res; } int main() { ll n, k; cin >> n >> k; vector l(n), r(n); rep(i, 0, n) cin >> l[i] >> r[i]; mint ans = 1; vector

p(n); rep(i, 0, n) p[i] = {l[i], r[i]}; sort(p.begin(), p.end()); BIT seg(200005); rep(i, 0, n) { auto [l, r] = p[i]; ans *= max(0, (int)k-seg.sum(l+1, 200005)); seg.add(r, 1); } cout << (mint)mod_pow(k, n) - ans << endl; return 0; }