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: Runtime Error)

#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;

typedef long long ll;

template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}

struct Work {
    int a; // days
    int b; // values
};

// const int MAX_N = 100000;
// const int MAX_M = 100000;
const int MAX_N = 10000;
const int MAX_M = 10000;

// int dp[MAX_N + 1][MAX_M + 1];

// int solve(const int N, const int M, const vector<struct Work> &works) {
//     for (int n = 1; n <= N; n++) {
//         for (int m = 1; m <= M; m++) {
//             if (m - works[n].a >= 0) {
//                 // chmax(dp[n][m], dp[n - 1][m - A[n]] + B[n]);
//                 chmax(dp[n][m], dp[n - 1][m - 1] + works[n].b);
//             }
//             chmax(dp[n][m], dp[n - 1][m]);
//         }
//     }
// 
//     return dp[N][M];
// }

int dp[MAX_M + 1];

int solve(const int N, const int M, const vector<struct Work> &works) {
    for (int n = 1; n <= N; n++) {
        for (int m = M; m >= 1; m--) {
            if (m - works[n].a >= 0) {
                // chmax(dp[n][m], dp[n - 1][m - A[n]] + B[n]);
                chmax(dp[m], dp[m - 1] + works[n].b);
            }
        }
    }

    return dp[M];
}

int main(void) {
    int N, M;
    cin >> N >> M;

    vector<struct Work> works(N + 1);
    for (int i = 1; i <= N; i++) {
        cin >> works[i].a >> works[i].b;
    }

    sort(works.begin(), works.end(), [](const struct Work &x, const struct Work &y) {
            return x.a < y.a;
            });

    cout << solve(N, M, works) << endl;

    return 0;
}