Time Travel and Multiset

Seonggwan built a time machine. This is an algorithm problem, so instead of doing something impressive with it he decided to apply time travel to a multiset.

A multiset with time travel supports three operations.

1. Go to time $t$ and put one copy of the integer $x$ into the multiset.
2. Go to time $t$ and take one copy of the integer $x$ out of the multiset. At least one copy of $x$ is guaranteed to be present at time $t$.
3. Print how many copies of the integer $x$ the multiset holds at time $t$.

An integer put in at time $t$ stays from time $t$ onward, and an integer taken out at time $t$ is gone from time $t$ onward. So the number of copies of $x$ at time $t$ is the number of already processed type 1 operations on $x$ whose time is at most $t$, minus the number of already processed type 2 operations on $x$ whose time is at most $t$.

For example, suppose two copies of the integer 1 are put in at time 2 and one copy is taken out at time 5. The integer 1 then has two copies from time 2 to time 4 and one copy from time 5 onward. If one more copy is taken out at time 4, the integer 1 has two copies from time 2 to time 3, one copy at time 4, and none from time 5 onward. In that state an operation that takes the integer 1 out at time 5 never appears in the input, because no copy exists at that moment.

Operations are processed in the order given in the input. The answer to a type 3 operation reflects only the type 1 and type 2 operations that come before it.

Write a program that behaves as described above.

The first line contains the number of queries $N$ ($1 \le N \le 100\,000$).

Each of the next $N$ lines contains one query as three positive integers $a_i$, $t_i$, $x_i$ ($1 \le a_i \le 3$, $1 \le t_i, x_i \le 10^9$). $a_i$ is the operation type, and $t_i$ and $x_i$ are the $t$ and $x$ from the statement.

For every query with $a_i = 3$, print its answer on its own line.