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.
Input
In the first line, there is an Integer T(1<=T<=10), which means the test cases in the input file. Then followed by T test cases.
For each test case, I will give you an Integer N(0<N<=100,000) in the first line, which means the number of stones on the road. Then followed by N lines and there are two integers Pi(0<=Pi<=100,000) and Di(0<=Di<=1,000) in the line, which means the position of the i-th stone and how far Sempr can throw it.
Output
Just output one line for one test case, as described in the Description.
Sample Input
2
2
1 5
2 4
2
1 5
6 6
Sample Output
11
12
/*题意：
题意：给定n个石头的位置pi,和能够扔的距离Di,从左(0位置)往右走，碰到的石头为奇数个就往右扔，碰到的石头为偶数个就跳过，问最后一个石头距离出发点的距离
直接优先队列模拟
思路：自定义优先级，按照总小到大的顺序出队，奇数时继续加入新的节点，直到队列为空判断结束，要用到pair不然有一个下标与扔出的距离的对应关系很难处理。
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <map>
using namespace std;
typedef pair<int,int>mp;
struct cmp{
bool operator () (const mp a, const mp b){
if(a.first == b.first)
return a.second > b.second;
else
return a.first > b.first;
}
};
int main()
{
int t;
scanf("%d", &t);
while(t--){
int n;
priority_queue <mp, vector<mp>, cmp> q;
scanf("%d", &n);
for(int i = 0; i < n; i++){
int x, l;
scanf("%d%d", &x, &l);
q.push(mp(x,l));
}
int sum = 0;
int maxn = 0;
int cnt = 1;
int j = 0;
while(!q.empty()){
int f = q.top().first;
int t = q.top().second;
int len = f + t; //因为接下来会将上一节点的信息pop掉所以先将上一节点信息保存在f，t中
q.pop();
if(cnt & 1){
q.push(mp(len, t));
maxn = max(maxn, len);
}
cnt++;
}
printf("%d\n", maxn);
}
return 0;
}
