Anonymity Is Important
Contest: Codeforces Round 773 (Div. 1)
Difficulty rating: 2200
Tags: binary search, brute force, data structures, dsu, greedy, sortings

In the work of a doctor, it is important to maintain the anonymity of clients and the results of tests. The test results are sent to everyone personally by email, but people are very impatient and they want to know the results right away.

That's why in the testing lab "De-vitro" doctors came up with an experimental way to report the results. Let's assume that $$$n$$$ people took the tests in the order of the queue. Then the chief doctor Sam can make several statements, in each telling if there is a sick person among the people in the queue from $$$l$$$-th to $$$r$$$-th (inclusive), for some values $$$l$$$ and $$$r$$$.

During the process, Sam will check how well this scheme works and will be interested in whether it is possible to find out the test result of $$$i$$$-th person from the information he announced. And if it can be done, then is that patient sick or not.

Help Sam to test his scheme.

Input format:
The first line contains two integers $$$n$$$, $$$q$$$ ($$$1 \le n, q \le 2 \cdot 10^5$$$) — the number of people and the number of queries.

In each of the next $$$q$$$ lines, the description of the query is given. The first number in the line is $$$t$$$ ($$$t = 0$$$ or $$$t = 1$$$) — the type of the query.

If $$$t = 0$$$, the line contains three more integers $$$l, r, x$$$ ($$$1 \le l \le r \le n$$$, $$$x = 0$$$ or $$$x = 1$$$). This query means that Sam tells that among the people in the queue from $$$l$$$-th to $$$r$$$-th (inclusive):

- there was at least one sick person, if $$$x=1$$$,
- there is no sick people, if $$$x=0$$$.

If $$$t = 1$$$, the line contains one more integer $$$j$$$ ($$$1 \le j \le n$$$) — the position of the patient in the queue, for which Sam wants to know the status.

All queries are correct, that means that there always exists an example of the queue of length $$$n$$$ for which all reported results (statements from queries with $$$t = 0$$$) are true.

Output format:
After each Sam question (query with $$$t = 1$$$) print:

- "NO", if the patient is definitely not sick,
- "YES", if the patient is definitely sick.
- "N/A", if it is impossible to definitely identify the status of patient having the given information.

Constraints:
Time limit: 1.5 s
Memory limit: 256.0 MB

Note:
In the first test for the five first queries:

1. Initially Sam tells that people $$$4$$$, $$$5$$$ are not sick.
2. In the next query Sam asks the status of the patient $$$5$$$. From the previous query, we know that the patient is definitely not sick.
3. In the next query Sam asks the status of the patient $$$6$$$. We don't know any information about that patient now.
4. After that Sam tells that there exists a sick patient among $$$4$$$, $$$5$$$, $$$6$$$.
5. In the next query Sam asks the status of the patient $$$6$$$. Now we can tell that this patient is definitely sick.

Example input:
6 9
0 4 5 0
1 5
1 6
0 4 6 1
1 6
0 2 5 1
0 2 2 0
1 3
1 2

Example output:
NO
N/A
YES
YES
NO

Editorial:
If $$$i$$$-th person is not ill, the following query exists: $$$0 \ l \ r \ 0$$$, such that $$$l \le i \le r$$$.
Otherwise, the person's status is either unknown or they are ill.
If $$$i$$$-th person is ill, the following query exists:$$$0 \ l \ r \ 1$$$, such that $$$l \le i \le r$$$, and every person $$$i$$$ such that $$$l \leq j \leq r$$$ are not ill. If there is such person $$$j$$$ that they are not ill, and $$$j \neq i, l \le j \le r$$$. In this case, it is impossible to determine if $$$i$$$-th person is ill or not.
Let's maintain the indices of the people who might be ill using set. When we get a query $$$0 \ l \ r \ 0$$$, we can find the first possible ill person with an index of at least $$$l$$$ using lower_bound, after that, we need to delete this person from our set, find the next one and do the same thing until we find the first index greater than $$$r$$$. This works in $$$O(nlogn)$$$.
If a person is not in the set, he is totally healthy. Otherwise, we can use a segment tree to store such index $$$j$$$ that there is a query $$$0 \ i \ j \ 1$$$ and store it in the $$$i$$$-th slot of our segment tree. We can update it when we get a new query. When we understand that the $$$i$$$-th person might be ill, we can find the first elements to the left ($$$l$$$) and to the right ($$$r$$$) of $$$i$$$, which might be ill using our set. The $$$i$$$-th person is ill when the minimal element on segment $$$[l + 1; i]$$$ is $$$< r $$$.
The solution works in $$$O(nlogn + qlogn)$$$.