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)

// ░░░░░░░░( •̪●)░░░░░░░░░░░░░░░░░░░░░░░░
// ░░░░░░███████ ]▄▄▄▄▄▄▄▄▃░░░▃░░░░ ▃░
// ▂▄▅█████████▅▄▃▂░░░░░░░░░░░░░░░░░
// [███████████████████].░░░░░░░░░░░░░░
// ◥⊙▲⊙▲⊙▲⊙▲⊙▲⊙▲⊙◤...░░░░░░░░░░░░

//PointBlank's code (⌐■_■)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
 
#define repp(I, A, B) for(int I = A; I <= B; I++)
#define rep(I, A, B) for(int I = A; I < B; I++)
#define rrep(I, B, A) for(int I = B; I >= A; I--)
#define pb emplace_back
#define ff first
#define ss second
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define LB lower_bound
#define UB upper_bound
#define MP make_pair
#define mem(A, B) memset(A, B, sizeof(A));
#define mem0(A) memset(A, 0, sizeof(A));
#define quickio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define debug(this) cerr<<"> "<<#this<<" : "<<this<<"\n"
#define bitsf(n) __builtin_popcount(n)
#define in_edges(M) repp(I, 1, M){ll A, B; cin >> A >> B; v[A].pb(B), v[B].pb(A);}
#define in_edges2(M) repp(I, 1, M){ll A, B; cin >> A >> B; v[A].pb(B);}
#define endl "\n"
ll power(ll x, ll y); 

const ll MOD = 1000000007;
 
int main() //PointBlank's code ¯\_(ツ)_/¯
{
	int n,m;
	cin >> n >> m;
	vector<int> v[m + 5];
	rep(i, 0, n){
		int a, b;
		cin >> a >> b;
		if(a > m) continue;
		v[m - a].pb(b);
	}
	priority_queue<int> q;
	ll res = 0;
	rrep(i, m - 1, 0){
		for(auto x : v[i]) q.push(x);
		if(!q.empty()) res += q.top(), q.pop();
	}

	cout << res;
}

ll power(ll x, ll y)
{ 
    ll res = 1; x %= MOD;
    while(y > 0){if(y & 1) res = (res*x) % MOD; y = y>>1, x = (x*x) % MOD;} 
    return res; 
}