Problem: Problem Statement

You are given a list of $N$ intervals. The $i$-th interval is $[l_i, r_i)$, which denotes a range of numbers greater than or equal to $l_i$ and strictly less than $r_i$. In this task, you consider the following two numbers:

* The minimum integer $x$ such that you can select $x$ intervals from the given $N$ intervals so that the union of the selected intervals is $[0, L)$.
* The minimum integer $y$ such that for all possible combinations of $y$ intervals from the given $N$ interval, it does cover $[0, L)$.



We ask you to write a program to compute these two numbers.

* * *

Input

The input consists of a single test case formatted as follows.

> $N$ $L$ $l_1$ $r_1$ $l_2$ $r_2$ $\vdots$ $l_N$ $r_N$

The first line contains two integers $N$ ($1 \leq N \leq 2 \times 10^5$) and $L$ ($1 \leq L \leq 10^{12}$), where $N$ is the number of intervals and $L$ is the length of range to be covered, respectively. The $i$-th of the following $N$ lines contains two integers $l_i$ and $r_i$ ($0 \leq l_i < r_i \leq L$), representing the range of the $i$-th interval $[l_i, r_i)$. You can assume that the union of all the $N$ intervals is $[0, L)$

Output

Output two integers $x$ and $y$ mentioned in the problem statement, separated by a single space, in a line.

Examples

Input| Output
---|---


3 3
0 2
1 3
1 2


|


2 3



2 4
0 4
0 4


|


1 1



5 4
0 2
2 4
0 3
1 3
3 4


|


2 4






Example

Input




Output

Solution (in CPP):

Solution:
#include <iostream>
#include <vector>
#include <queue>
#include <math.h>
#include <set>
#include <random>
#include <map>
#include <unordered_map>
#include <bits/stdc++.h>
#define FOR(i, n, m) for(ll i = n; i < (int)m; i++)
#define REP(i, n) FOR(i, 0, n)
#define ALL(v) v.begin(), v.end()
#define pb push_back
using namespace std;
using ll = std::int_fast64_t;
using P = pair<ll, ll>;
constexpr ll inf = 1000000000;
constexpr ll mod = 1000000007;
constexpr long double eps = 1e-15;

template<typename T1, typename T2>
ostream& operator<<(ostream& os, pair<T1, T2> p) {
    os << to_string(p.first) << " " << to_string(p.second);
    return os;
}
template<typename T>
ostream& operator<<(ostream& os, vector<T>& v) {
    REP(i, v.size()) {
        if(i) os << " ";
        os << to_string(v[i]);
    }
    return os;
}

struct modint {
    ll n;
public:
    modint(const ll n = 0) : n((n % mod + mod) % mod) {}
    static modint pow(modint a, int m) {
        modint r = 1;
        while(m > 0) {
            if(m & 1) { r *= a; }
            a = (a * a); m /= 2;
        }
        return r;
    }
    modint &operator++() { *this += 1; return *this; }
    modint &operator--() { *this -= 1; return *this; }
    modint operator++(int) { modint ret = *this; *this += 1; return ret; }
    modint operator--(int) { modint ret = *this; *this -= 1; return ret; }
    modint operator~() const { return (this -> pow(n, mod - 2)); } // inverse
    friend bool operator==(const modint& lhs, const modint& rhs) {
        return lhs.n == rhs.n;
    }
    friend bool operator<(const modint& lhs, const modint& rhs) {
        return lhs.n < rhs.n;
    }
    friend bool operator>(const modint& lhs, const modint& rhs) {
        return lhs.n > rhs.n;
    }
    friend modint &operator+=(modint& lhs, const modint& rhs) {
        lhs.n += rhs.n;
        if (lhs.n >= mod) lhs.n -= mod;
        return lhs;
    }
    friend modint &operator-=(modint& lhs, const modint& rhs) {
        lhs.n -= rhs.n;
        if (lhs.n < 0) lhs.n += mod;
        return lhs;
    }
    friend modint &operator*=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * rhs.n) % mod;
        return lhs;
    }
    friend modint &operator/=(modint& lhs, const modint& rhs) {
        lhs.n = (lhs.n * (~rhs).n) % mod;
        return lhs;
    }
    friend modint operator+(const modint& lhs, const modint& rhs) {
        return modint(lhs.n + rhs.n);
    }
    friend modint operator-(const modint& lhs, const modint& rhs) {
        return modint(lhs.n - rhs.n);
    }
    friend modint operator*(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * rhs.n);
    }
    friend modint operator/(const modint& lhs, const modint& rhs) {
        return modint(lhs.n * (~rhs).n);
    }
};
istream& operator>>(istream& is, modint m) { is >> m.n; return is; }
ostream& operator<<(ostream& os, modint m) { os << m.n; return os; }

#define MAX_N 3030303
long long extgcd(long long a, long long b, long long& x, long long& y) {
    long long d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1; y = 0;
    }
    return d;
}
long long mod_inverse(long long a, long long m) {
    long long x, y;
    if(extgcd(a, m, x, y) == 1) return (m + x % m) % m;
    else return -1;
}
vector<long long> fact(MAX_N+1, inf);
long long mod_fact(long long n, long long& e) {
    if(fact[0] == inf) {
        fact[0]=1;
        if(MAX_N != 0) fact[1]=1;
        for(ll i = 2; i <= MAX_N; ++i) {
            fact[i] = (fact[i-1] * i) % mod;
        }
    }
    e = 0;
    if(n == 0) return 1;
    long long res = mod_fact(n / mod, e);
    e += n / mod;
    if((n / mod) % 2 != 0) return (res * (mod - fact[n % mod])) % mod;
    return (res * fact[n % mod]) % mod;
}
// return nCk
long long mod_comb(long long n, long long k) {
    if(n < 0 || k < 0 || n < k) return 0;
    long long e1, e2, e3;
    long long a1 = mod_fact(n, e1), a2 = mod_fact(k, e2), a3 = mod_fact(n - k, e3);
    if(e1 > e2 + e3) return 0;
    return (a1 * mod_inverse((a2 * a3) % mod, mod)) % mod;
}

using mi = modint;

mi mod_pow(ll a, ll n) {
    mi ret = 1;
    mi tmp = a;
    while(n > 0) {
        if(n % 2) ret *= tmp;
        tmp = tmp * tmp;
        n /= 2;
    }
    return ret;
}

template <typename T> struct segment_tree {
private:
    int n;
    const function<T(T, T)> op;
    const T ie;
    vector<T> seq;
    
public:
    segment_tree(int _n, function<T(T, T)> op, const T ie) : op(op), ie(ie) {
        n = 1;
        while(n < _n) n *= 2;
        seq.resize(2 * n - 1);
        for(int i = 0; i < 2 * n - 1; i++) seq[i] = ie;
    }
    
    void update(int k, const T e) {
        k += n - 1;
        seq[k] = e;
        while(k > 0) {
            k = (k - 1) / 2;
            seq[k] = op(seq[k * 2 + 1], seq[k * 2 + 2]);
        }
    }
    
    T get(int k) {
        k += n - 1;
        return seq[k];
    }
    
    T query(int a, int b, int k = 0, int l = 0, int r = -1) {
        if(r == -1) r = n;
        if(r <= a || b <= l) return ie;
        if(a <= l && r <= b) return seq[k];
        T vl = query(a, b, k * 2 + 1, l, (l + r) / 2);
        T vr = query(a, b, k * 2 + 2, (l + r) / 2, r);
        return op(vl, vr);
    }
};
// segment_tree<ll> rsq(n, [] (ll a, ll b) { return a + b; }, 0)
// segment_tree<ll> rMq(n, [] (ll a, ll b) { return max(a, b); }, -1e18)
// segment_tree<ll> rmq(n, [] (ll a, ll b) { return min(a, b); }, 1e18)


ll N, L;
vector<ll> l(202020), r(202020);

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> N >> L;
    REP(i, N) cin >> l[i] >> r[i];
    set<ll> s;
    unordered_map<ll, int> m;
    REP(i, N) {
        s.insert(l[i]);
        s.insert(r[i]);
    }
    int cnt = 0;
    for(auto elm: s) {
        m[elm] = cnt;
        cnt++;
    }
    segment_tree<ll> rMq(cnt + 1, [] (ll a, ll b) { return max(a, b); }, -1e18);
    REP(i, N) {
        rMq.update(m[l[i]], max(r[i], rMq.get(m[l[i]])));
    }
    vector<int> imos(cnt, 0);
    REP(i, N) {
        imos[m[l[i]]]++;
        imos[m[r[i]]]--;
    }
    FOR(i, 1, cnt) imos[i] += imos[i - 1];
    int ans1 = 0, ans2 = inf;
    ll tmp = 0;
    while(tmp != L) {
        tmp = rMq.query(0, m[tmp] + 1);
        ans1++;
    }
    REP(i, cnt - 1) ans2 = min(ans2, imos[i]);
    cout << ans1 << " " << N - ans2 + 1 << endl;
    return 0;
}