Pseudocode Task

Hint: This code calculates a sum based on nested loops, breaking out of the inner loop under a specific condition. Consider how the break statement affects the final values of `count` and `accumulator`.

count = 0
accumulator = 0
FOR x IN RANGE(1, 4): // Python's range is exclusive for the end value
    FOR y IN RANGE(1, 5): // Python's range is exclusive for the end value
        IF x == 2 AND y == 3:
            BREAK // Exits the innermost loop (y-loop)
        END IF
        accumulator = accumulator + y
        count = count + 1
    END FOR // y-loop
END FOR // x-loop
PRINT count
PRINT accumulator