B. Integers Shop-Hello 2022

Problem - 1621B - Codeforces

B. Integers Shop

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The integers shop sells nn segments. The ii-th of them contains all integers from lili to riri and costs cici coins.

Tomorrow Vasya will go to this shop and will buy some segments there. He will get all integers that appear in at least one of bought segments. The total cost of the purchase is the sum of costs of all segments in it.

After shopping, Vasya will get some more integers as a gift. He will get integer xx as a gift if and only if all of the following conditions are satisfied:

• Vasya hasn't bought xx.
• Vasya has bought integer ll that is less than xx.
• Vasya has bought integer rr that is greater than xx.

Vasya can get integer xx as a gift only once so he won't have the same integers after receiving a gift.

For example, if Vasya buys segment [2,4][2,4] for 2020 coins and segment [7,8][7,8] for 2222 coins, he spends 4242 coins and receives integers 2,3,4,7,82,3,4,7,8 from these segments. He also gets integers 55 and 66 as a gift.

Due to the technical issues only the first ss segments (that is, segments [l1,r1],[l2,r2],…,[ls,rs][l1,r1],[l2,r2],…,[ls,rs]) will be available tomorrow in the shop.

Vasya wants to get (to buy or to get as a gift) as many integers as possible. If he can do this in differents ways, he selects the cheapest of them.

For each ss from 11 to nn, find how many coins will Vasya spend if only the first ss segments will be available.

Input

The first line contains a single integer tt (1≤t≤10001≤t≤1000) — the number of test cases.

The first line of each test case contains the single integer nn (1≤n≤1051≤n≤105) — the number of segments in the shop.

Each of next nn lines contains three integers lili, riri, cici (1≤li≤ri≤109,1≤ci≤1091≤li≤ri≤109,1≤ci≤109) — the ends of the ii-th segments and its cost.

It is guaranteed that the total sum of nn over all test cases doesn't exceed 2⋅1052⋅105.

Output

For each test case output nn integers: the ss-th (1≤s≤n1≤s≤n) of them should be the number of coins Vasia will spend in the shop if only the first ss segments will be available.

Example

input

Copy

3
2
2 4 20
7 8 22
2
5 11 42
5 11 42
6
1 4 4
5 8 9
7 8 7
2 10 252
1 11 271
1 10 1

output

Copy

20
42
42
42
4
13
11
256
271
271

Note

In the first test case if s=1s=1 then Vasya can buy only the segment [2,4][2,4] for 2020 coins and get 33 integers.

The way to get 77 integers for 4242 coins in case s=2s=2 is described in the statement.

In the second test case note, that there can be the same segments in the shop.

=========================================================================

First of all, make sure that when the number is the most , Two paragraphs of numbers must be chosen , And no choice , After the selection, there is no need to re select , So we maintain left,right The extreme value of . Once you can modify , Modify immediately left, And its cost , And it should be revised together with the answer . If left Same as current , Then take the minimum , Revise with the answer .

The special case is , What we are currently encountering is the maximum interval , Then we have to modify the left and right expenses , But we can change the answer to the minimum cost per time and itself .

# include<iostream>
# include<algorithm>
# include<math.h>
using namespace std;
typedef long long int ll;

ll lef[200000+10],righ[200000+10],cost[200000+10];


int main()
{


    /*
         When the left endpoint is modified but the right endpoint is not 

         The left endpoint cost must be modified , The answer should also be modified 

         The same is true for the right endpoint 

         When the left endpoint is the same as the original left endpoint   Change the left endpoint cost to a smaller value 

         The same is true for the right endpoint 

         Special judgement 
         When the left end point and the right end point cover all the original sections , The answer should be output ,


    */

    int t;

    cin>>t;


    while(t--)
    {

        int n;

        cin>>n;


        for(int i=1;i<=n;i++)
        {
            cin>>lef[i]>>righ[i]>>cost[i];
        }

        ll ans=0x7f7f7f7f7f7f7f7f;

        ll leftans=ans,rightans=ans;

        ll l=ans,r=0;


        for(int i=1;i<=n;i++)
        {

            if(l>lef[i])
            {
                l=lef[i];

                leftans=cost[i];

                ans=leftans+rightans;


            }
            else if(l==lef[i])
            {

                leftans=min(leftans,cost[i]);

                ans=min(ans,leftans+rightans);



            }

            if(r<righ[i])
            {
                r=righ[i];

                rightans=cost[i];


                ans=leftans+rightans;

            }
            else if(r==righ[i])
            {
               rightans=min(rightans,cost[i]);


               ans=min(ans,leftans+rightans);


            }

            if(r==righ[i]&&l==lef[i])
            {
                ans=min(ans,cost[i]);

            }


            cout<<ans<<endl;


        }

    }

    return 0;

}