Problem:
Score : 400 points

Problem Statement

You are going out for a walk, when you suddenly encounter a monster. Fortunately, you have N katana (swords), Katana 1, Katana 2, …, Katana N, and can perform the following two kinds of attacks in any order:

Wield one of the katana you have. When you wield Katana i (1 ≤ i ≤ N), the monster receives a_i points of damage. The same katana can be wielded any number of times.

Throw one of the katana you have. When you throw Katana i (1 ≤ i ≤ N) at the monster, it receives b_i points of damage, and you lose the katana. That is, you can no longer wield or throw that katana.

The monster will vanish when the total damage it has received is H points or more. At least how many attacks do you need in order to vanish it in total?

Constraints

1 ≤ N ≤ 10^5

1 ≤ H ≤ 10^9

1 ≤ a_i ≤ b_i ≤ 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N H
a_1 b_1
:
a_N b_N

Output

Print the minimum total number of attacks required to vanish the monster.

Sample Input 1

1 10
3 5

Sample Output 1

3

You have one katana. Wielding it deals 3 points of damage, and throwing it deals 5 points of damage. By wielding it twice and then throwing it, you will deal 3 + 3 + 5 = 11 points of damage in a total of three attacks, vanishing the monster.

Sample Input 2

2 10
3 5
2 6

Sample Output 2

2

In addition to the katana above, you also have another katana. Wielding it deals 2 points of damage, and throwing it deals 6 points of damage. By throwing both katana, you will deal 5 + 6 = 11 points of damage in two attacks, vanishing the monster.

Sample Input 3

4 1000000000
1 1
1 10000000
1 30000000
1 99999999

Sample Output 3

860000004

Sample Input 4

5 500
35 44
28 83
46 62
31 79
40 43

Sample Output 4

9

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <iomanip>
#include <bitset>
#include <set>
#include <map>
#include <stdio.h>
#include <numeric>
#include <cstring>

#define rep(i,n) for (int i = 0; i < (n); i++)
#define rep1(i,n) for (int i = 1; i < (n); i++)
#define FOR(i,a,b) for (int i=(a); i < (b); i++)
#define MOD 1000000007 //10^9+7
#define ALL(a)  (a).begin(),(a).end()
using namespace std;
using ll = long long;
using PII = pair<int, int>;
using PLL = pair<long long, long long>;
const int INF = numeric_limits<int>::max();
const ll INFL = numeric_limits<ll>::max();
constexpr ll TEN(int n) { return (n==0) ? 1 : 10*TEN(n-1); }

// 負の数にも対応した % 演算
long long mod(long long val, long long m) {
    long long res = val % m;
    if (res < 0) res += m;
    return res;
}

//greatest common divisor
long long gcd(ll a, ll b) 
{
    if (a % b == 0) {
        return b;
    } else {
        return gcd(b, a % b);
    }
}

//least common multiple
long long lcm(ll a, ll b)
{
    return a / gcd(a, b) * b ;
}

ll factorial(ll n) {
    ll res = 1;
    for (ll i = 1; i <= n; i++) {
        res *= i;
    }
    return res;
}

bool fn_is_prime(ll n) {
    for (ll i = 2; i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return n != 1;
}

bool isOK(int index, int key, vector<ll>& a) {
    if (a[index] >= key) return true;
    else return false;
}

int binary_search(int key, vector<ll>& a) {
    int ng = -1;
    int ok = a.size();

    while (abs(ok - ng) > 1) {
        int mid = (ok + ng) / 2;

        if (isOK(mid, key, a)) ok = mid;
        else ng = mid;
    }
    return ok;
}

struct UnionFind {
    vector<int> r;

    UnionFind(int N) {
        r = vector<int>(N, -1);
    }

    int root(int x) {
        if (r[x] < 0) return x;
        return r[x] = root(r[x]);
    }

    bool unite(int x, int y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;
        if (r[x] > r[y]) swap(x, y);
        r[x] += r[y];
        r[y] = x;
        return true;
    }

    int size(int x) {
        return -r[root(x)];
    }

    bool same(int x, int y) {
        return root(x) == root(y);
    }
};

//素因数分解
vector<pair<long long, long long> > prime_factorize(long long N) {
    vector<pair<long long, long long> > res;
    for (long long a = 2; a * a <= N; ++a) {
        if (N % a != 0) continue;
        long long ex = 0;
        while (N % a == 0) {
            ++ex;
            N /= a;
        }
        res.push_back({a, ex});
    }
    if (N != 1) res.push_back({N, 1});
    return res;
}
//const auto &pf = prime_factorize(i);
//for (auto p : pf) cout << p.first << " " << p.second;

int main()
{
    ll n, h; cin >>n >> h;
    vector<PLL> k(n);
    ll amax = 0;
    rep(i,n) {
        ll a, b; cin >> a >> b;
        k[i] = PLL{b, a};
        amax = max(amax, a);
    }

    sort(k.begin(), k.end());
    reverse(ALL(k));
    ll ans = 0;
    int index = 0;
    rep(i,n) {
        if (k[i].first < amax)
            break;
        h -= k[i].first;
        ans++;
        if (h <= 0) {
            cout << ans << endl;
            return 0;
        }
    }
    if (h % amax == 0)
        ans += h/amax;
    else
        ans += h/amax + 1;
    cout << ans << endl;

    return 0;
}