A cinema has `n` rows of seats, numbered from 1 to `n` and there are ten seats in each row, labelled from 1 to 10 as shown in the figure above.

Given the array `reservedSeats` containing the numbers of seats already reserved, for example, `reservedSeats[i] = [3,8]` means the seat located in row **3** and labelled with **8** is already reserved.

_Return the maximum number of four-person groups you can assign on the cinema seats._ A four-person group occupies four adjacent seats **in one single row**. Seats across an aisle (such as \[3,3\] and \[3,4\]) are not considered to be adjacent, but there is an exceptional case on which an aisle split a four-person group, in that case, the aisle split a four-person group in the middle, which means to have two people on each side.

**Example 1:**

**Input:** n = 3, reservedSeats = \[\[1,2\],\[1,3\],\[1,8\],\[2,6\],\[3,1\],\[3,10\]\]
**Output:** 4
**Explanation:** The figure above shows the optimal allocation for four groups, where seats mark with blue are already reserved and contiguous seats mark with orange are for one group.

**Example 2:**

**Input:** n = 2, reservedSeats = \[\[2,1\],\[1,8\],\[2,6\]\]
**Output:** 2

**Example 3:**

**Input:** n = 4, reservedSeats = \[\[4,3\],\[1,4\],\[4,6\],\[1,7\]\]
**Output:** 4

**Constraints:**

*   `1 <= n <= 10^9`
*   `1 <= reservedSeats.length <= min(10*n, 10^4)`
*   `reservedSeats[i].length == 2`
*   `1 <= reservedSeats[i][0] <= n`
*   `1 <= reservedSeats[i][1] <= 10`
*   All `reservedSeats[i]` are distinct.


def shiftGrid(grid, k):
    m, n = len(grid), len(grid[0])
    k %= (m * n)
    res = [[0] * n for _ in range(m)]
    for i in range(m):
        for j in range(n):
            idx = (i * n + j + k) % (m * n)
            res[idx // n][idx % n] = grid[i][j]
    return res
