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)

//============================================================================
//  Falling Down is an accident, Staying Down is a Choice.
// Author      : Murad
// Online Judge: Codeforces.cpp & Atcoder.cpp
// Description : Problem name
//============================================================================
#include<bits/stdc++.h>
#include<unordered_map>
#include<unordered_set>
#include<utility>
using namespace std;
typedef                                    long long LL;
#define                                    pi 3.1415926536
#define                                    forn(i,a,b)for(int i=a;i<b;i++)
#define                                    ULL unsigned long long
#define                                    MP make_pair
#define                                    ff first
#define                                    ss second
#define                                    endl '\n'
#define                                    INF LL(1e9);
#define                                    _INF INT32_MIN
#define                                    pq priority_queue
#define                                    MM multimap
#define                                    PB push_back
#define                                    EMPP emplace_back
#define                                    vii vector<int>
#define                                    vll vector<LL>
#define                                    ipair pair<int,int>
#define                                    lpair pair<LL,LL>
#define                                    clr(v, d) memset(v, d, sizeof(v))
#define                                    El3zba ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define                                    modd 1000000007
#define                                    sf1(v) scanf("%I64d",&v);
#define                                    sf2(v1,v2) scanf("%I64d %I64d",&v1,&v2)
#define                                    sf3(v1,v2,v3) scanf("%I64d %I64d %I64d",&v1,&v2,&v3)
                                           //std::transform(s1.begin(), s1.end(), s1.begin(),::tolower);
                                           char  alphz[27] = { 'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z' };
float Euclidean(LL x1, LL x2, LL y1, LL y2) {
	return sqrt(pow(x1 - x2, 2) + pow(y1 - y2, 2));
}
LL GCD(LL a, LL b) {
	return !b ? a : GCD(b, a % b);
}
LL LCM(LL a, LL b) {
	return (a*b) / GCD(a, b);
}
void PrimeFactor(LL n)
{
	while (n % 2 == 0) {
		//printf("%d ", 2);
//		aa.insert(2);
		n /= 2;
	}
	for (int i = 3; i <= sqrt(n); i += 2)
	{
		if (n%i == 0) {
			//printf("%d ", i);
			//aa.insert(i);
			n /= i;
		}
	}
	if (n > 2) {
		//printf("%d ", n);
	//	aa.insert(n);
	};
}
bool Is_Square(LL x) {
	LL l = 0, r = x;
	while (l <= r) {
		LL mid = l + (r - l) / 2;
		if (mid*mid == x)return true;
		if (mid*mid > x)r = mid - 1;
		else l = mid + 1;
	}
	return false;
}
LL Power(LL x, LL y)
{
	LL temp;
	if (y == 0)return 1;
	temp = Power(x, y / 2);
	if (y % 2 == 0)
		return (temp*temp);
	else return (x*temp*temp);
}
bool Is_Prime(int x) {
	if (x == 2)return 1;
	else if (x % 2 == 0 || x < 2)return 0;
	for (int i = 3; i*i <= x; i += 2)
		if (x%i == 0)return 0;
	return 1;
}
bool Is_Palin(string s)
{
	int i = 0, j = (int)s.size() - 1;
	while (i < j)
	{
		if (s[i] != s[j])return 0;
		i++, j--;
	}
	return 1;
}
const int N = 10005;
int mem[N][N];
vii a, b;
int solv(int idx, int m,int n) {
	if (idx >= n)
		return 0;
	if (mem[idx][m] != -1)
		return mem[idx][m];
	int go1 = 0, go2 = 0, go3 = 0;
	if (idx + a[idx] <= m) {
		 go1 = solv(idx + 1, m, n);
		 go2 = solv(idx + 1, m, n) + b[idx];
	}
	if(idx+a[idx]>m) go3 = solv(idx + 1, m, n);
	mem[idx][m] = max({ go1,go2,go3 });
	return mem[idx][m];
}
int main()
{
	El3zba;
	/***  بسم الله الرحمن الرحيم   ***/
	int n, m;
	cin >> n >> m;
	a.resize(n);
	b.resize(n);
	clr(mem, -1);
	for (int i = 0; i < n; i++)
		cin >> a[i] >> b[i];
	int ans = solv(0, m, n);
	cout << ans << endl;
	return 0;
}