The traveler’s position after any number of moves is constrained on a fixed circle which is the circle that passes through the three points: $$$P$$$, position after first move, position after second move. For the traveler to come back to $$$P$$$, he will have to head in the same direction as the initial direction and if he is heading in the same direction as the initial direction then, he must be at $$$P$$$(else his subsequent positions wont lie on the same circle). Thus, $$$n\cdot a$$$ must be a multiple of $$$360$$$. The first time this happens is when he turns a total angle of $$$lcm(a,360)$$$, i.e, the number of moves is $$$\frac{lcm(a,360)}{a}$$$. Thus the total distance is $$$d \cdot \frac{lcm(a,360)}{a}$$$.

Let the direction along which the traveler moves at $$$P$$$ be the $$$y$$$-axis and $$$P$$$ be the origin. Suppose after $$$n$$$ moves, the person is at $$$(x,y)$$$. Then, $$$x=\sum^{n-1}_{i=0}d\cdot \sin(i\cdot a)=\frac{d\cdot \sin(\frac{n\cdot a}{2})\sin(\frac{(n-1)a}{2})}{\sin(\frac{a}{2})}$$$ and $$$y=\sum^{n-1}_{i=0}d\cdot \cos(i\cdot a)=\frac{d\cdot \sin(\frac{n\cdot a}{2})\cos(\frac{(n-1)a}{2})}{\sin(\frac{a}{2})}$$$.

For both the $$$x$$$ and $$$y$$$ positions to be simultaneously $$$0$$$, $$$\frac{n\cdot a}{2}$$$ must be a multiple of $$$180$$$ thus, $$$n\cdot a$$$ must be a multiple of $$$360$$$.

The first time this happens is when total angle turned equals $$$lcm(a,360)$$$, i.e, the number of moves is $$$\frac{lcm(a,360)}{a}$$$. Thus the total distance is $$$d\cdot \frac{lcm(a,360)}{a}$$$.

Note that if $$$n=1$$$, then Alice always wins, since Alice has 1 point and Bob has 0.

Suppose $$$n \gt 1$$$, we can greedily assume that Alice and Bob always pick a number that gives $$$n$$$ points whenever possible. Thus, if there are $$$k$$$ numbers such that $$$gcd(i,n) = 1$$$, then Alice picks $$$\lceil \frac{k}{2} \rceil$$$ such numbers and Bob picks $$$\lfloor \frac{k}{2} \rfloor$$$ such numbers. Hence, if $$$k$$$ is even, then Alice wins if $$$n$$$ is odd and it is a tie if $$$n$$$ is even. However, if $$$k$$$ is odd, then Alice always wins.

Turns out, for $$$n \gt 2$$$, $$$k$$$ is always even. This can be showed in multiple ways, such as the formula for the Euler Totient function. One other method is to observe that $$$gcd(i, n) = gcd(n-i, n)$$$. This implies that there always an even number of $$$i$$$ such that $$$gcd(i,n) = 1$$$ (Note that $$$i=n-i$$$ implies $$$i = \frac{n}{2}$$$, and thus $$$gcd(i, n) \neq 1$$$.).

Since we're given that $$$a_i \leq 2^{15}-1$$$, all $$$a_i$$$'s are 15 bit numbers. Thus, if $$$n \gt 15$$$, we can set the $$$i$$$-th bit in $$$a_i$$$ by performing the operation on $$$a_i$$$ and $$$a_{16}$$$ if the bit is 0. Thus in this case, the answer is always $$$2^{15}-1$$$.

Now for $$$n \leq 15$$$, we can simply brute force what elements should be flipped. We just need to ensure that the total number of elements whose bits are flipped is even.

Claim: For $$$n \geq 8$$$, answer is always $$$2^{63}-1$$$.

Proof: You need to set a total of 63 bits in order to reach $$$2^{63}-1$$$. We can try to set half of the required bits in each step as follows:

- Look at $$$a_1$$$. If less than 32 bits are set in $$$a_1$$$, then flip $$$a_1$$$ along with $$$a_8$$$. Otherwise, do nothing. Now, we've set at least 32 of the required 63 bits. So remaining number of bits we should set is at most 31.

- Look at $$$a_2$$$. If less than 16 of the remaining bits (the ones which were not set in $$$a_1$$$) are set in $$$a_2$$$, then flip $$$a_2$$$ and $$$a_8$$$. Otherwise, do nothing. Now we've set atleast 32 + 16 = 48 bits.

We can continue on in this manner until we are left with no more bits to set. This takes at most $$$\lceil logn \rceil$$$ steps, and we can reach an optimal solution in under 7 operations. We can simply brute force for $$$n \leq 7$$$.

Note that this means we can solve the problem in at most 4 operations as well. We can eliminate all numbers of the array except the first 8, and then brute force on this array. Since a bitmask of size 8 can have at most $$$4$$$ pairs of $$$1$$$, we can solve it in 4 operations.

Let us fix the number of non-zero elements in the final array. Let us call this number $$$M$$$ ($$$0 \leq M \leq n)$$$. We will try to solve this problem for this fixed $$$M$$$.

Notice that the minimum sum that you can construct with $$$M$$$ non-zero elements is by choosing the first $$$M$$$ elements of the array. The maximum sum is by choosing the largest $$$M$$$ elements.

Claim: We can make any target sum which lies between the minimum and maximum sum using these operations.

Proof: Let us start by choosing the smallest $$$M$$$ elements of the array, say $$$a_1, a_2, \ldots, a_m$$$. If the sum of these elements is less than $$$k$$$ (our required sum), then we can choose $$$a_1, a_2, \ldots, a_{m-1}, a_{m+1}$$$ and increase our current sum by 1. We can keep on moving the last element until it is $$$a_n$$$. Suppose our current sum is still less than $$$k$$$, then we start moving $$$a_{m-1}$$$ towards the end one step at a time. We can keep on repeating this process for $$$a_{m-2}$$$, $$$a_{m-3}$$$, $$$\ldots$$$ until we reach our target sum. Since we always increase our current sum by $$$1$$$, we are guaranteed to reach every possible sum between the maximum and minimum.

Notice that during this process, the number of subarrays that we set to zero is at most $$$2$$$ at every step. Thus, we can guarantee that the answer will be $$$\leq 2$$$ if it is possible. We can brute force check for answer = $$$0,1$$$. If we don't find a valid solution, then we iterate over $$$M$$$, and check if $$$k$$$ lies between the corresponding minimum and maximum possible sum. If it does, then we can find the answer for this $$$M$$$. If no such $$$M$$$ exists, then output -1, as we have checked all possible cases of $$$M$$$.