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 FOR(i,n) for(int i=0;i<n;i++)
#define why(n,x) int n;while(cin >>n,n!=x)
#define iFOR(i,x,n)for(int i=x;i<n;i++)
#define fin <<'\n'
#define __ <<" "<<
#define ___ <<" "
#define bash push_back
#define ALL(x) x.begin(),x.end()
#define SWAP(a,b) ((a != b) && (a += b,b = a - b,a -= b))
//#define int long long
typedef long long ll;
typedef pair < int , int >  pii;
typedef vector < int > vit;
typedef map < string , int > mstit;
typedef vector < pii > vpi ;
typedef greater < pii > gpi;
typedef priority_queue < pii , vpi , gpi > dijk ;
static const signed int INF = 0x3f3f3f3f;
static const signed long long LINF = 0x3f3f3f3f3f3f3f3fLL;
static const signed int SMOD = 1000000007;
static const signed int NMOD = 1000000009;
static const signed int dx[]={1,0,-1,0,1,1,-1,-1};
static const signed int dy[]={0,-1,0,1,-1,1,-1,1};

bool inside(int x,int y,int w,int h){return (x>=0 && y>=0 && x<w && y<h);}
template<class T>T abs(T &x){return x<0 ? -x : x;}
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
int qp(int a,ll b,int mo){int ans=1;do{if(b&1)ans=1ll*ans*a%mo;a=1ll*a*a%mo;}while(b>>=1);return ans;}
int gcd(int a,int b){return b?gcd(b,a%b):a;}

int n, m;
vector<int> a, b, c[100005];

priority_queue<int> que;
inline void solve(){

    cin >> n >> m;

    a.resize(n+5);
    b.resize(n+5);
    for(int i = 0; i < n; i++) {
        cin >> a[i] >> b[i];
        c[a[i]].bash(b[i]);
        
    }

    ll ans = 0;
    
    for(int i = 1; i <= m; i++) {
        for(auto&&x : c[i])
            que.push(x);
        if(!que.empty()){
            ans += que.top();
            que.pop();
        }
    }

    cout << ans << endl;
    
}
struct xyz{ xyz(){ cin.tie(0), ios::sync_with_stdio(false); cout <<fixed<<setprecision(12);};}xyzzy;signed main(){solve();return 0;}