Because of the wrong status of the bicycle, Sempr begin to walk east to west every morning and walk back every evening. Walking may cause a little tired, so Sempr always play some games this time. 
There are many stones on the road, when he meet a stone, he will throw it ahead as far as possible if it is the odd stone he meet, or leave it where it was if it is the even stone. Now give you some informations about the stones on the road, you are to tell me the distance from the start point to the farthest stone after Sempr walk by. Please pay attention that if two or more stones stay at the same position, you will meet the larger one(the one with the smallest Di, as described in the Input) first. 

2
2
1 5
2 4
2
1 5
6 6

11
12

优先队列，一次放入，定义优先级位置靠前距离小的石头优先级高，奇数个取出栈顶更新后放入，偶数个直接取出，队列为空结束。

#include<stdio.h>
#include<queue>
using namespace std;
struct Node{
    int x,y;
    friend bool operator<(Node a,Node b)
    {
        if(a.x==b.x) return a.y>b.y;
        return a.x>b.x;
    }
}node,no;

int main()
{
    int t,n,c,max,i;
    priority_queue<Node> q;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        for(i=1;i<=n;i++){
            scanf("%d%d",&node.x,&node.y);
            q.push(node);
        }
        c=0;max=0;
        while(q.size()){
            c++;
            no.x=q.top().x,no.y=q.top().y;
            q.pop();
            if(c%2==1){
                no.x+=no.y;
                if(no.x>max) max=no.x;
                q.push(no);
            }
        }
        printf("%d\n",max);
    }    
    return 0;
}