读入之后换不同的预处理方式会被不同的点卡。。。思路是分开处理a=0的挂件和a>=2的挂件，如何花愉悦值购买a>=2的挂件。

#include <bits/stdc++.h>

#define N 2010
#define mp(a,b) make_pair(a,b)
#define endl '\n'
#define int long long

using namespace std ; 

typedef pair<int,int> pii ;
 
int dp[N][N<<1] , s[N] , pri[N] , n ; 

const int inf = (1ll<<62) ; 
#define fi first
#define se second
#define pb push_back 

vector < int > v0 ;  
vector < pii > v2 ; 

signed main () {
	
	cin.sync_with_stdio(false) ; 
	
	cin >> n ; 
	int ans = 0 , sta = 1 ; 
	/*
	用这个WA第10个点 
	for ( int i = 1 , a , b ; i <= n ; i++ ) {
		cin >> a >> b ; 
		if ( a == 1 && (b >  0 ) ) ans += b ; // ok
		else if ( a == 0 && (b >= 0 ) ) v0.push_back(b) ; //	???
		else if ( a >= 2 && (b >= 0 ) ) sta += a-1 , ans += b ; 
		else if ( a >= 2 && (b <  0 ) ) v2.push_back({a-1,-b}) ; 
		else if ( (a <= 1) && (b <= 0) ) continue ;  
	}
	*/
	
	/*
	用这个WA第1个点 
	for ( int i = 1 , a , b ; i <= n ; i++ ) {
			cin >> a >> b ; 
			if ( a == 1 && (b > 0 ) ) ans += b ; // ok ,  ????
			else if ( a == 0 ) v0.push_back(b) ; // ??? a == 0 , b <= 0 ?  a == 1 , b <= 0 ?
			else if ( a >= 2 && b >= 0 ) sta += a-1 , ans += b ; 
			else if ( a >= 2 && b < 0 )  v2.push_back({a-1,-b}) ; 
		}
	*/
	
	sort    (v0.begin(),v0.end()) ; 
	reverse (v0.begin(),v0.end()) ; 
	
	for ( int i = 1 ; i <= v0.size() ; i++ ) 
		s[i] = s[i-1]+v0[i-1] ;
	
	for ( int j = 1 ; j <= n+1 ; j++ ) pri[j] = inf ; 
	sort   (v2.begin(),v2.end()) ; 
	reverse(v2.begin(),v2.end()) ;  
	
	for ( int i = 1 ; i <= v2.size() ; i++ ) {
		if ( v0.size() - v2[i-1].fi >= 0 ) pri[v0.size()] = min ( pri[ v0.size() - v2[i-1].fi ] + v2[i-1].se , pri[v0.size()] ) ;
		else pri[v0.size()] = min ( pri[0] + v2[i-1].se , pri[v0.size()] ) ;
		
		for ( int jj = v0.size()-v2[i-1].fi ; jj >= 0 ; jj-- ) {
			int j = jj+v2[i-1].fi ;
			pri[j] = min( pri[j] , pri[jj] + v2[i-1].se )  ; 
		}
		
		for ( int j = v0.size(); j >= 0 ; j-- ) 
			pri[j] = min( pri[j+1] , pri[j] ) ; 
	}
	
	
	int sum = s[sta] ; 
	
	if ( sta >= v0.size() ) sum = s[v0.size()] ; 
	
	
	for ( int i = sta ; i <= v0.size() ; i++ ) {
		int now = s[i] - pri[i-sta] ; 
		sum = max(sum,now) ; 
	}
	
	ans += sum ; 
	
	cout << ans << endl ; 
	
	return 0 ; 
}