A

Farmer John's N (1 <= N <= 50,000) cows (numbered 1..N) are planning to run away and join the circus. Their hoofed feet prevent them from tightrope walking and swinging from the trapeze (and their last attempt at firing a cow out of a cannon met with a dismal failure). Thus, they have decided to practice performing acrobatic stunts. 

The cows aren't terribly creative and have only come up with one acrobatic stunt: standing on top of each other to form a vertical stack of some height. The cows are trying to figure out the order in which they should arrange themselves ithin this stack. 

Each of the N cows has an associated weight (1 <= W_i <= 10,000) and strength (1 <= S_i <= 1,000,000,000). The risk of a cow collapsing is equal to the combined weight of all cows on top of her (not including her own weight, of course) minus her strength (so that a stronger cow has a lower risk). Your task is to determine an ordering of the cows that minimizes the greatest risk of collapse for any of the cows.

Input

* Line 1: A single line with the integer N. 

* Lines 2..N+1: Line i+1 describes cow i with two space-separated integers, W_i and S_i. 

Output

* Line 1: A single integer, giving the largest risk of all the cows in any optimal ordering that minimizes the risk.

Sample Input

310 32 53 3

Sample Output

2

Hint

OUTPUT DETAILS: 

Put the cow with weight 10 on the bottom. She will carry the other two cows, so the risk of her collapsing is 2+3-3=2. The other cows have lower risk of collapsing.

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iomanip>
using namespace std;
#define inf 0x3f3f3f
struct node
{
    int w,s;
};
bool cmp(node a,node b)
{
    return (a.w+a.s)<(b.w+b.s);
}
int main()
{
    int n,i,max;
    long long sum[50010];
    node a[50010];
    while(cin>>n)
    {
        for(i=1;i<=n;i++)
        //cin>>a[i].w>>a[i].s;
        scanf("%d%d",&a[i].w,&a[i].s);
        memset(sum,0,sizeof(sum));
        sort(a+1,a+1+n,cmp);
        for(i=1;i<=n;i++)
         sum[i]=sum[i-1]+a[i].w;
        //cout<<sum[n-1]-a[n].s<<endl;
        max=-inf;
        for(i=n;i>=1;i--)//一开始写的i>=2;最上面的一头牛也得看。。
        {
            if(max<sum[i-1]-a[i].s)
                max=sum[i-1]-a[i].s;
        }
        cout<<max<<endl;
    }
}
//ps:要求的是对于牛的各种排序在每一种排序中选出每头牛倒塌风险最大的，然后计算最大值得最小是多少，一开始以为二分，但是不能分，没有定值
//然后想按重量大的在下面排序，这样风险就小，但是这是两个都有影响，应该是二者之和最大的在下面，在比赛过程中如果没思路这样的题可以猜猜按什么排序
//无非加减乘除