超市裡有n個產品要賣，每個產品都有一個截至時間dx（從開始賣時算起），只有在這個截至時間之前才能賣出並且獲得率潤dy。 有多個產品，所有可以有不同的賣出順序，每賣一個產品要佔用1個單位的時間，問最多能賣出多少利潤。 　　 A supermarket has a set Prod of products on sale. It earns a profit px for each product x∈Prod sold by a deadline dx that is measured as an integral number of time units starting from the moment the sale begins. Each product takes precisely one unit of time for being sold. A selling schedule is an ordered subset of products Sell ≤ Prod such that the selling of each product x∈Sell, according to the ordering of Sell, completes before the deadline dx or just when dx expires. The profit of the selling schedule is Profit(Sell)=Σ_x∈Sellpx. An optimal selling schedule is a schedule with a maximum profit.
For example, consider the products Prod={a,b,c,d} with (pa,da)=(50,2), (pb,db)=(10,1), (pc,dc)=(20,2), and (pd,dd)=(30,1). The possible selling schedules are listed in table 1. For instance, the schedule Sell={d,a} shows that the selling of product d starts at time 0 and ends at time 1, while the selling of product a starts at time 1 and ends at time 2. Each of these products is sold by its deadline. Sell is the optimal schedule and its profit is 80.

Write a program that reads sets of products from an input text file and computes the profit of an optimal selling schedule for each set of products.

Input

A set of products starts with an integer 0 <= n <= 10000, which is the number of products in the set, and continues with n pairs pi di of integers, 1 <= pi <= 10000 and 1 <= di <= 10000, that designate the profit and the selling deadline of the i-th product. White spaces can occur freely in input. Input data terminate with an end of file and are guaranteed correct.

Output

For each set of products, the program prints on the standard output the profit of an optimal selling schedule for the set. Each result is printed from the beginning of a separate line.

Sample Output

80
185

Hint

The sample input contains two product sets. The first set encodes the products from table 1. The second set is for 7 products. The profit of an optimal schedule for these products is 185.

#include <iostream>
#include 
 
<string.h>
#include <queue>
#include <algorithm>
using namespace std;
struct node
{
    int p,d;
};
node q[10010];
int pre[10010];
int cmp(node a,node b)
{
    return a.p>b.p;
}
int find(int x)
{
    return x==pre[x]?x:find(pre[x]);
}
int main()
{
    int n;
    while(cin>>n)
    {
        for(int i=1;i<=10010;i++) pre[i]=i;
        for(int i=1;i<=n;i++)
        cin>>q[i].p>>q[i].d;
        sort(q+1,q+1+n,cmp);//商品從價格從大到小排序
        int sum=0;
        for(int i=1;i<=n;i++)
        {
            int k=find(q[i].d);//找到第i個最大的價格的天數的祖宗，及找到最後期限，然後日期給k
            if(k>0)//如果在天數為0之前，那麼
            {
                sum+=q[i].p;//總利潤
                pre[k]=k-1;
　　　　　　　　/*他的根節點，也就是賣出去的日期數就要-1, 這樣，
　　　　　　　　　下次遇到相同結束日期的商品的時候，它的根節點的值， 
　　　　　　　　　也就是賣出去的日期數就是現在的k-1*/
            }
        }
        cout<<sum<<endl;
    }
}