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++ (GCC 9.2.1) (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<vector>
#include<functional>
#include<algorithm>
#include<stdlib.h>
#include<string>
#include<string.h>
#define _USE_MATH_DEFINES
#include<math.h>
#include<deque>
#include<set>
#include<map>
#include<queue>
#include<list>
#include<iostream>
#include <bitset>
using namespace std;
typedef long long ll;
#define rep(i,a,b) for(auto i=a;i<b;i++)
#define rep2(i, a)for(auto i : a)
#define all(_x) _x.begin(), _x.end()
#define r_sort(_x) sort(_x, std::greater<int>())
#define vec_cnt(_a, _n) (upper_bound(all(_a), _n) - lower_bound(all(_a), _n))
#define vec_unique(_a) _a.erase(std::unique(all(_a)), _a.end());
#define vvec vector<vector<ll>>
ll gcd(ll a, ll b) { return a % b == 0 ? b : gcd(b, a % b); }
ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; }
#define INF 1 << 30
const int mod = 1000000007;
ll power(ll x, ll p) { ll a = 1; while (p > 0) { if (p % 2 == 0) { x *= x; p /= 2; } else { a *= x; p--; } }return a; }
ll mpower(ll x, ll p) { ll a = 1; while (p > 0) { if (p % 2 == 0) { x = x * x % mod; p /= 2; } else { a = a * x % mod; p--; } }return a; }
ll ac(ll n, ll k) { ll a = 1; rep(i, 1, k) { a *= n - i + 1; a /= i; }return a; }
ll mc(ll n, ll m) { ll k = 1, l = 1; rep(i, n - m + 1, n + 1) k = k * i % mod; rep(i, 1, m + 1) l = l * i % mod; l = mpower(l, mod - 2); return k * l % mod; }

int main()
{
	int n, m, a, b;
	cin >> n >> m;
	vector<vector<int>> work(m + 1);
	rep(i, 0, n) {
		cin >> a >> b;
		if (a > m)continue;
		work[a].push_back(b);
	}
	priority_queue<int> temp;
	ll ans = 0;
	rep(i, 0, m) {
		rep2(v, work[i + 1])temp.push(v);
		if (temp.size()) {
			ans += (ll)temp.top();
			temp.pop();
		}
	}
	printf("%lld\n", ans);
	return 0;
}