This is the easy version of the problem. The difference between the versions is that in this version, the time limit and the constraints on $n$ and $q$ are lower. You can hack only if you solved all versions of this problem.

Gellyfish has an array consisting of $n$ sets. Initially, all the sets are empty.

Now Gellyfish will do $q$ operations. Each operation contains one modification operation and one query operation, for the $i$-th ($1 \leq i \leq q$) operation:

First, there will be a modification operation, which is one of the following:

1. Insert operation: You are given an integer $r$. For the $1$-th to $r$-th sets, insert element $i$. Note that the element inserted here is $i$, the index of the operation, not the index of the set.
2. Reverse operation: You are given an integer $r$. Reverse the $1$-th to $r$-th sets.
3. Delete operation: You are given an integer $x$. Delete element $x$ from all sets that contain $x$.

Followed by a query operation:

• Query operation: You are given an integer $p$. Output the smallest element in the $p$-th set (If the $p$-th set is empty, the answer is considered to be $0$).

Now, Flower needs to provide the answer for each query operation. Please help her!

Additional constraint on the problem: Gellyfish will only give the next operation after Flower has answered the previous query operation. That is, you need to solve this problem online. Please refer to the input format for more details.

Input

The first line contains two integers $n$ and $q$ ($1 \leq n, q \leq 10^5$) — the number of the sets and the number of operations.

As you need to respond to the operations online, the operations will be encoded.

The $i$-th line of the following $q$ lines contains three integers $a$, $b$, and $c$ ($1 \leq a \leq 3$, $1 \leq c \leq n$) — describing the $i$-th operation in an encoded form.

Here, $a$ represents the type of modification operation. Among them, $a=1$ represents Insert operation, $a=2$ represents Reverse operation, $a=3$ represents Delete operation.

• If $a = 1$, then the modification operation is the Insert operation. It will be guaranteed that $1 \leq b \leq n$. $r$ will be calculated as $r=(b+\text{ans}_{i-1}-1) \bmod n + 1$.
• If $a=2$, then the modification operation is the Reverse operation. It will be guaranteed that $1 \leq b \leq n$. $r$ will be calculated as $r=(b+\text{ans}_{i-1}-1) \bmod n + 1$.
• If $a=3$, then the modification operation is the Delete operation. It will be guaranteed that $1 \leq b \leq q$. $x$ will be calculated as $x=(b+\text{ans}_{i-1}-1) \bmod q + 1$.

For the query operation, $p$ will be calculated as $p = (c+\text{ans}_{i-1}-1) \bmod n + 1$.

Here $ \text{ans}_{i} (1 \leq i \leq q)$ represents the answer to the query operation in the $i$-th operation. Additionally, we define $ \text{ans}_{0} = 0$.

Output

For each query operation, output the answer to the query.

Note

All the sets are empty in the beginning, so the array is $[\{\}, \{\}, \{\}, \{\}, \{\}]$.

With the decoding method given before, we can see what happens in each operation:

1. For the first operation: $a = 1, r = 2, p = 2$. The modification operation is an Insert operation; element $1$ is inserted into the first two sets; so the array becomes $[\{1\}, \{1\}, \{\}, \{\}, \{\}]$, and the smallest element in the second set is $1$.
2. For the second operation: $a = 2, r = 4, p = 2$. The modification operation is a Reverse operation; the first four sets are reversed; so the array becomes $[\{\}, \{\}, \{1\}, \{1\}, \{\}]$, and the second set is empty, which means the answer is $0$.
3. For the third operation: $a = 1, r = 5, p = 3$. The modification operation is an Insert operation; element $3$ is inserted into all the sets; so the array becomes $[\{3\}, \{3\}, \{1, 3\}, \{1, 3\}, \{3\}]$, and the smallest element in the third set is $1$.
4. For the fourth operation: $a = 2, r = 3, p = 1$. The modification operation is a Reverse operation; the first three sets are reversed; so the array becomes $[\{1, 3\}, \{3\}, \{3\}, \{1, 3\}, \{3\}]$, and the smallest element in the first set is $1$.
5. For the fifth operation: $a = 1, r = 1, p = 3$. The modification operation is an Insert operation; element $5$ is inserted into the first set; so the array becomes $[\{1, 3, 5\}, \{3\}, \{3\}, \{1, 3\}, \{3\}]$, and the smallest element in the third set is $3$.
6. For the sixth operation: $a = 2, r = 2, p = 2$. The modification operation is a Reverse operation; the first two sets are reversed; so the array becomes $[\{3\}, \{1, 3, 5\}, \{3\}, \{1, 3\}, \{3\}]$, and the smallest element in the second set is $1$.
7. For the seventh operation: $a = 3, x = 3, p = 3$. The modification operation is a Delete operation; element $3$ is deleted from all the sets; so the array becomes $[\{\}, \{1, 5\}, \{\}, \{1\}, \{\}]$, and the third set is empty, which means the answer is $0$.
8. For the eighth operation: $a = 3, x = 1, p = 2$. The modification operation is a Delete operation; element $1$ is deleted from all the sets; so the array becomes $[\{\}, \{5\}, \{\}, \{\}, \{\}]$, and the smallest element in the second set is $5$.
9. For the ninth operation: $a = 3, x = 5, p = 5$. The modification operation is a Delete operation; element $5$ is deleted from all the sets; so the array becomes $[\{\}, \{\}, \{\}, \{\}, \{\}]$, and the fifth set is empty, which means the answer is $0$.
10. For the tenth operation: $a = 3, x = 2, p = 4$. The modification operation is a Delete operation; element $2$ is deleted from all the sets; so the array becomes $[\{\}, \{\}, \{\}, \{\}, \{\}]$, and the fourth set is empty, which means the answer is $0$.

Please note that although we have not inserted element $2$ into the sets, we still delete element $2$ from all the sets in the tenth operation, which means that the Delete operation doesn't necessarily require the existence of a set to contain the deleted element. It also shows that it is possible to have two Delete operations that delete the same element.