Ok so let's have the following cards (A and B respectively)

4 6

8 8

1 9

And queries 2 8 9. Now the idea is to find for each card the last query that was in the interval formed A and B (including A, excluding B).

For card 1, there is no such query.as there is no query of value 4 or 5.

For card 2, there is no such query too, since there is no interval when A=B

For card 3 the last query is the second, since the first and second are in the inteval (1 to 8) and the third isn't.

Finding those values is done with dynamic rmq over all 10^9 possible values of T. So in this case the values we need to update are v[2]=1, v[8]=2, v[9]=3. For example for the third card, the answer is the biggest number in interval A~B-1 (i.e. 1 to 8).

Now knowing those we sort by the value we just found, having the largest numbers before the smallest. We get the sequence of card 3,1,2. Now we keep another segment tree over 10^9 leaves. Each time we process a card, we add all queries that weren't prevously added and are larger than the value we found for this card. Then we calculate other values.for the cards. Let's see it as an example.

First we process card 3. The last query that fits in its interval is the second. Since the third is not added, we add it. That means val[9]++ in our segment tree. Then the answer we are looking for, call it flips[] is simply a query from B to infinity (or max value, i.e. 10^9). So flips[3]=Sum(9,inf), and since only val[9] is 1, answer is 1. We get flips[3]=1.

When processing card 1 we will add queries 1 and 2 since no queries fitted in its interval, therefore the last query to fit is said to be 0, i.e. val[8]++ and val[2]++. Then we get flips[1]=Sum(6,inf) = 2.

Lastly processig card 2, we've already added all queries, so flips[2]=Sum(8,inf)=2

Now let's see all the information we have. For the first card, it started at the smaller face and there were no queries in its interval. Therefore all flips of that card are equal to flips[1]=2. Therefore this card faced A at the end, i.e. 4.

The second card also didn't have a query in its interval. It also had flips[2]=2, so it faced side A at the end, and that is 8.

The third card again started while.facing the smaller number, but however did have queries in its interval, last of them — the second query. After the second query was done, we can be sure that this card was facing the bigger face, i.e. B. Now the flips that were done AFTER that query are saved in flips[3]=1. So after being at B e flip once and face A. Then the third card again show face A, i.e. 1.

The total sum is 4+8+1=13.

The answer is 13.

Feel free to tell me if you still don't understand.