Mental Arithmetic Board Sum

The jury chair of a mental arithmetic championship came up with a new task. Initially $n$ integers $a_1, a_2, \ldots, a_n$ are written on the board. After that, the contestant must execute commands of two types:

1. Erase the $i$-th number and write $x$ in its place. The numbers become $a_1, \ldots, a_{i-1}, x, a_{i+1}, \ldots, a_n$.
2. Cyclically shift the sequence to the right by $k$. The numbers become $a_{n-k+1}, \ldots, a_n, a_1, \ldots, a_{n-k}$.

After each command, the contestant must compute the sum of all numbers on the board and report it to the jury. The jury members must compute the required sums themselves to check the answers.

The first line contains an integer $n$, the number of integers initially on the board ($2 \leq n \leq 10^5$).

The second line contains $n$ integers separated by spaces: $a_1, a_2, \ldots, a_n$ ($-10^9 \leq a_i \leq 10^9$).

The third line contains an integer $q$, the number of commands ($1 \leq q \leq 10^5$).

Each of the next $q$ lines contains one command in one of the following formats:

• 1 i x: replace the $i$-th number with $x$ ($1 \leq i \leq n$; $-10^9 \leq x \leq 10^9$).
• 2 k: cyclically shift the sequence to the right by $k$ ($1 \leq k < n$).

Print $q$ lines, each containing one integer. The $i$-th line is the sum of the numbers on the board after the first $i$ commands.

The answer can be large, so a 64-bit integer type is needed: int64 in Pascal, long long in C++, long in Java.

In the example from the statement, the initial sequence is $4, 1, 2, 1, 5, 3$. After the first command, the sequence is shifted right by 3 elements, giving $1, 5, 3, 4, 1, 2$ with sum $1 + 5 + 3 + 4 + 1 + 2 = 16$. The second command replaces the third element with $10$, giving $1, 5, 10, 4, 1, 2$ with sum $23$. The third command replaces the fourth element with $4$. It is already $4$, so the sequence does not change and the sum stays $23$. The fourth command shifts the sequence right by 1, giving $2, 1, 5, 10, 4, 1$. The sum does not change. The fifth command replaces the first element with $-10$, giving $-10, 1, 5, 10, 4, 1$ with sum $-10 + 1 + 5 + 10 + 4 + 1 = 11$.