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 (Clang 3.8.0) (judge verdict: Accepted)

#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>
#include <string>
#include <bitset>
#include <queue>
#include <tuple>
#include <cmath>
#include <map>
#include <cstring>

template<typename T> bool chmax( T &a, const T &b ) { if ( a <= b ) { a = b; return ( true ); } else { return ( false ); } }
template<typename T> bool chmin( T &a, const T &b ) { if ( a >= b ) { a = b; return ( true ); } else { return ( false ); } }

using namespace std;

using ll = long long;
using ull = unsigned long long;

using Pint = pair<int, int>;
using Pll  = pair<ll, ll>;
using Pull = pair<ull, ull>;

#define eb emplace_back
#define pb push_back
#define mp make_pair
#define mt make_tuple

#define F first
#define S second

#define rep( i, n ) for ( int i = 0; i < (int)( n ); ++i )
#define reps( i, n ) for ( int i = 1; i <= (int)( n ); ++i )
#define rrep( i, n ) for ( int i = (int)( ( n ) - 1 ); i >= 0; --i )
#define rreps( i, n ) for ( int i = (int)( ( n ) ); i > 0; --i )
#define arep( i, v ) for ( auto &&i : ( v ) )

template<typename T> T gcd( const T a, const T b ) { return ( b ? gcd( b, a % b ) : a ); }
template<typename T> T lcm( const T a, const T b ) { return ( a / gcd( a, b ) * b ); }

#define ALL( c ) ( c ).begin(), ( c ).end()
#define RALL( c ) ( c ).rbegin(), ( c ).rend()
#define UNIQUE( c ) ( c ).erase( unique( ( c ).begin(), ( c ).end() ), ( c ).end() )

constexpr ll MOD = 1000000007LL;
template<typename T = ll> constexpr T MAX = numeric_limits<T>::max();

#define y0 y3487465
#define y1 y8687969
#define j0 j1347829
#define j1 j234892
#define next asdnext
#define prev asdprev

template<typename T = ll> class UnionFind {
public:
	UnionFind( T n ) { par.resize( n ); siz.resize( n ); rep( i, n ) { par[i] = i; siz[i] = 1; } }
	T find( T x ) { if ( x == par[x] ) return ( x ); else return( par[x] = find( par[x] ) ); }
	void unite( T x, T y ) { T xx = find( x ); T yy = find( y ); if ( xx == yy ) return;
		if ( siz[xx] <= siz[yy] ) swap( xx, yy ); par[yy] = xx; siz[xx] += siz[yy]; }
private:
	vector<T> par, siz;
};

template<typename T = ll> class CompareMax {
public:	T operator()( T a, T b ) { return ( max( a, b ) ); }
};
template<typename T = ll> class CompareMin {
public:	T operator()( T a, T b ) { return ( min( a, b ) ); }
};

template<typename T = ll, typename F = CompareMax<T>, T I = 0> class SegTree {
public:
	SegTree( T n ) { N = n; v.resize( 4 * n ); rep( i, 4 * n ) v[i] = I; }
	void update( T i, T x ) { i += N - 1; v[i] = x; while ( i > 0 ) {
			i = ( i - 1 ) / 2; v[i] = func( v[i * 2 + 1], v[i * 2 + 2] ); } }
	T query( T a, T b, T k, T l, T r ) {
		if ( r <= a || b <= l ) return ( I );
		if ( a <= l && r <= b ) return ( v[k] );
		else { 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 ( func( vl, vr ) ); }
	}
private:
	T N; F func; vector<T> v;
};

template<typename T = ll> T solveLIS( const vector<T> &v ) {
	vector<T> dp( v.size(), numeric_limits<T>::max() );
	rep( i, v.size() ) {
		*lower_bound( ALL( dp ), v[i] ) = v[i];
	}
	return ( distance( dp.begin(), lower_bound( ALL( dp ), numeric_limits<T>::max() ) ) );
}

template<typename T = ll> T power( T a, T b, T m = MOD ) {
	T res = 1;
	while ( b > 0 ) { if ( b & 1 ) res = res * a % m;
		a = a * a % m; b >>= 1; }
	return ( res );
}

/*
constexpr ll COMBSIZE = 500010;
ll fact[COMBSIZE];
ll inv[COMBSIZE];
ll inv_fact[COMBSIZE];

template<typename T> void initComb( T n, T m = MOD )
{
	fact[0] = fact[1] = inv_fact[0] = inv_fact[1] = 1;
	inv[1] = 1;
	for ( int i = 2; i < n; i++ ) {
		fact[i] = ( fact[i - 1] * i ) % m;
		inv[i] = m - inv[m % i] * ( m / i ) % m;
		inv_fact[i] = inv_fact[i - 1] * inv[i] % m;
	}
}

template<typename T> T comb( T n, T r, T m = MOD )
{
	if ( n < r ) return ( 0 );
	if ( n < 0 || r < 0 ) return ( 0 );
	return ( fact[n] * ( inv_fact[r] * inv_fact[n - r] % m ) % m );
}
*/

void replace( string &s, string t, string r ) {
	string::size_type p = 0;
	while ( ( p = s.find( t, p ) ) != string::npos ) {
		s.replace( p, t.length(), r );
		p += r.length();
	}
}

int main()
{
	ll N, M;
	cin >> N >> M;
	map<ll, vector<ll>> m;
	rep( i, N )
	{
		ll a, b;
		cin >> a >> b;
		m[a].eb( b );
	}

	ll ans = 0;
	priority_queue<ll> q;
	reps( i, M )
	{
		if ( m[i].empty() == false )
		{
			for ( auto &&it : m[i] )
				q.emplace( it );
		}

		if ( q.empty() == false )
		{
			ll tmp = q.top();
			q.pop();
			ans += tmp;
		}
	}

	cout << ans << endl;

	return ( 0 );
}