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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
using ll=long long;
using ld=long double;
using pll=pair<ll,ll>;
using plll=pair<ll,pll>;
using graph=vector<vector<int>>;

const ll MOD=1000000007;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};

ll abs1(ll a){
  if(a>=0)return a;
  else return -a;
}

ll pow1(ll a,ll b,ll p=LLONG_MAX){
  ll ret=1;
  while(b!=0){
    if((b&1)==1){
      ret*=a;
    }
    a*=a;
    if(a>=p)a%=p;
    b=b>>1;
  }
  return ret;
}

int digit(ll a,ll b){
  int i=0;
  while(a!=0){
    a/=b;
    i++;
  }
  return i;
}

ll gcd1(ll a,ll b){
  if(a==0 && b==0)return 0;
  ll n;
  while(a%b!=0){
    n=a%b;
    a=b;
    b=n;
  }
  return b;
}

ll solveLinear(ll n,ll m,ll c){
  if(n==0 && m==0)return 0;
  ll a[100],b[100];
  a[0]=n;b[0]=m;
  int i=1;
  while(b[i-1]!=0){
    a[i]=b[i-1];
    b[i]=a[i-1]%b[i-1];
    i++;
  }
  n=i;
  if(c%a[n-1]!=0){
    return 0;
  }
  ll x=c/a[n-1],y=0;
  for(i=1;i<n;i++){
    ll cnt=x;
    x=y;
    y=cnt-(a[n-i-1]/b[n-i-1])*y;
  }
  return x;
}

ll modInv(ll a,ll p=MOD){
  a%=p;
  ll n=solveLinear(a,p,1);
  n%=p;
  if(n<0)return n+p;
  return n;
}

ll fact(ll n,ll p=MOD){
  if(n<0)return 0;
  int i;
  ll ret=1;
  for(i=0;i<n;i++){
    ret=ret*(i+1);
    if(ret>=p)ret%=p;
  }
  return ret;
}

ll perm(ll n,ll k,ll p=MOD){
  if(n<0 || k<0 || n<k)return 0;
  int i;
  ll ret=1;
  for(i=0;i<k;i++){
    ret=ret*(n-i);
    if(ret>=p)ret%=p;
  }
  return ret;
}

ll binom(ll n,ll k,ll p=MOD){
  if(n<0 || k<0 || n<k)return 0;
  int i;
  ll ret=1,fact=1;
  for(i=0;i<k;i++){
    ret=ret*(n-i);
    fact*=i+1;
    if(ret>=p)ret%=p;
    if(fact>=p)fact%=p;
  }
  return (ret*modInv(fact))%p;
}

void pushHeap(vector<int>& h,int n){
  int i=h.size();
  h.push_back(n);
  while(i!=0 && h[(i-1)/2]<h[i]){
    swap(h[i],h[(i-1)/2]);
    i=(i-1)/2;
  }
  h[i]=n;
}

int popHeap(vector<int>& h){
  int n=h.size()-1;
  if(n==-1)return 0;
  int i,ret=h[0];
  swap(h[0],h[n]);
  h.pop_back();
  for(i=0;i<n;){
    int j=2*i+1;
    if(j>=n)break;
    if(j!=n-1 && h[j]<h[j+1])j++;
    if(h[i]<h[j]){
      swap(h[i],h[j]);
    }
    i=j;
  }
  return ret;
}

vector<int> ver;
void dfs(const graph &g,int v){
  ver[v]=1;
  for(auto i:g[v]){
    if(ver[i]!=1)continue;
    dfs(g,i);
  }
}

int main(){
  int i,j,k,cnt=0;
  int n,m;
  cin>>n>>m;
  vector<pll> v;
  for(i=0;i<n;i++){
    int a,b;
    cin>>a>>b;
    v.push_back(pll(b,a));
  }
  sort(v.begin(),v.end());
  ll ans=0;
  int d[m+1];
  for(i=0;i<=m;i++)d[i]=0;
  while(!v.empty()){
    int n=v.size()-1;
    int j=v[n].second;
    while(j<=m && d[j]>=v[n].first)j++;
    if(j<=m)d[j]=v[n].first;
    v.pop_back();
  }
  for(i=1;i<=m;i++)ans+=d[i];
  cout<<ans<<endl;
}