Problem:
Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

Take and complete the first job today. You will earn the reward of 3 after four days from today.

Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

Sample Input 2

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

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define NDEBUG
#include <cassert>


typedef long long ll;
typedef long double Double;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef pair<ll,ll> llll;
typedef pair<double,double> dd;

typedef vector<int> vi;
typedef vector<vector<int>> vvi;
typedef vector<ii> vii;
typedef vector<vector<ii>> vvii;
typedef vector<ll> vll;
typedef vector<vector<ll>> vvll;
typedef vector<llll> vllll;
typedef vector<bool> vb;
typedef vector<string> vs;
typedef vector<double> vd;
typedef vector<long double> vD;

#define sz(a)  int((a).size())
#define pb  push_back
#define eb  emplace_back
#define FOR(var,from,to) for(int var=(from);var<=(to);++var)
#define rep(var,n)  for(int var=0;var<(n);++var)
#define rep1(var,n)  for(int var=1;var<=(n);++var)
#define repC2(vari,varj,n)  for(int vari=0;vari<(n)-1;++vari)for(int varj=vari+1;varj<(n);++varj)
#define repC3(vari,varj,vark,n)  for(int vari=0;vari<(n)-2;++vari)for(int varj=vari+1;varj<(n)-1;++varj)for(int vark=varj+1;vark<(n);++vark)
#define ALL(c)  (c).begin(),(c).end()
#define RALL(c)  (c).rbegin(),(c).rend()
#define tr(i,c)  for(auto i=(c).begin(); i!=(c).end(); ++i)
#define mset(arr,val)  memset(arr,val,sizeof(arr))
#define mid(x,y) ((x)+((y)-(x))/2)
#define IN(x,a,b) ((a)<=(x)&&(x)<=(b))
#define cons make_pair
#define clamp(v,lo,hi) min(max(v,lo),hi)

template <typename T>
inline bool found(set<T> &s, T elem) { return s.find(elem) != s.end(); }
template <typename T>
inline bool found(unordered_set<T> &s, T elem) { return s.find(elem) != s.end(); }
template <typename T, typename U>
inline bool found(map<T,U> &s, U elem) { return s.find(elem) != s.end(); }
template <typename T, typename U>
inline bool found(unordered_map<T,U> &s, U elem) { return s.find(elem) != s.end(); }
template <typename T>
inline bool found(vector<T> &s, T elem) { return find(s.begin(), s.end(), elem) != s.end(); }
inline int found(string &s, string t) { return (int)s.find(t); }
inline int found(string &s, int c) { return (int)s.find(c); }

template<typename T1, typename T2> inline void amin(T1 & a, T2 const & b) { if (a>b) a=b; }
template<typename T1, typename T2> inline void amax(T1 & a, T2 const & b) { if (a<b) a=b; }
template<typename X, typename T> auto vectors(X x, T a) { return vector<T>(x, a); }
template<typename X, typename Y, typename Z, typename... Zs> auto vectors(X x, Y y, Z z, Zs... zs) { auto cont = vectors(y, z, zs...); return vector<decltype(cont)>(x, cont); }

inline ll square(ll x) { return x * x; }
inline ll gcd(ll a, ll b) { while(a) swap(a, b%=a); return b; }
template <typename T>
inline T mod(T a, T b) { return ((a % b) + b) % b; }

template <typename T>
int find_left(vector<T>& v, T elem) {
    return (int)(upper_bound(v.begin(), v.end(), elem) - v.begin()) - 1;
}
template <typename T>
int find_right(vector<T>& v, T elem) {
    return (int)(lower_bound(v.begin(), v.end(), elem) - v.begin());
}

ll solve(int N,int M, vi& a,vi& b) {
    priority_queue<ii> sch;
    rep(i,N){
        int d=M-a[i];
        if (d >= 0){
            sch.emplace(d, b[i]);
        }
    }

    ll ans =0;
    priority_queue<int> pq;
    for (int day=M-1; day>=0; --day) {
        while (!sch.empty()) {
            int d = sch.top().first, y = sch.top().second;
            if (d < day) break;
            sch.pop();
            pq.push(y);
        }
        if (pq.empty()) {
            continue;
        }
        {
            int y = pq.top(); pq.pop();
            ans += y;
        }
    }
    return ans;
}

int main() {
    int N, M; scanf("%d %d%*c", &N, &M);
    vi a(N),b(N); rep(i,N) scanf("%d %d%*c", &a[i], &b[i]);
    cout << solve(N,M,a,b) << endl;
    return 0;
}