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: Wrong Answer)

#include <iostream>
#include <iomanip>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <cstdio>
#include <utility>
#include <string>
#include <cmath>
#include <cstdlib>
#include <cstring>

using namespace std;

typedef uint64_t u64;
typedef int64_t s64;
typedef uint32_t u32;
typedef int32_t s32;
typedef vector<s32> vs32;
typedef vector<u32> vu32;
typedef vector<s64> vs64;
typedef vector<u64> vu64;

const double PI=3.14159265358979323846;

#define MAX(x, y) ((x) < (y) ? (y) : (x))
#define MIN(x, y) ((x) > (y) ? (y) : (x))

#define rep(i, N) for(int i = 0; i < N; ++i)

#define CEIL(x, y) (((x) + (y) - 1) / (y))
#define MOD 1000000007ULL

vector<pair<int, int>> daily;
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N, M;
    int A, B;
    cin >> N >> M;
    daily.resize(N);

    rep (i, N)
    {
        cin >> A >> B;
        daily[i].first = A;
        daily[i].second = B;
    }

    auto cmp = [](pair<int, int>& x, pair<int, int>& y) { return x.first > y.first; };
    sort(daily.begin(), daily.end(), cmp);

    u64 ret = 0;
    auto left = daily.begin();
    for (int i = 0; i < MIN(M, N); ++i)
    {
        auto it = lower_bound(left, daily.end(), pair<int, int>(M - i, 0), [](const pair<int, int>& x, const pair<int, int>& y) {return x.first > y.first; });
        if (it == daily.end()) break;
        left = it;
        it = max_element(it, daily.end(), [](const pair<int, int>& x, const pair<int, int>& y) {return x.second < y.second; });
        ret += it->second;
        daily.erase(it);
    }

    cout << ret << "\n";

    return 0;
}