Split

The word subsequence is mostly decoration here. Since $f$ only cares about occurrence counts, the order of elements does not matter. We are just deciding, for each distinct value, how many copies go into $p$. Then $q$ gets the rest.

Contribution of one value

Fix some value $x$ with total count $c$. Suppose $p$ gets $k$ copies of it, so $q$ gets $c-k$ copies.

This value contributes:

• $1$ to $f(p)$ if $k$ is odd,
• $1$ to $f(q)$ if $c-k$ is odd.

Now split by the parity of $c$.

If $c$ is odd, then $k$ and $c-k$ have opposite parity. Exactly one of them is odd, always. So every odd-frequency value contributes exactly $1$, no matter how we split it. There is nothing to optimize there.

If $c$ is even, then $k$ and $c-k$ have the same parity. So this value contributes:

• $2$ if $k$ is odd,
• $0$ if $k$ is even.

So even-frequency values are the only interesting ones. For each such value, we want to put an odd number of copies into $p$.

Let:

• $O$ = number of values with odd total frequency,
• $E$ = number of values with positive even total frequency.

The $O$ part is fixed. The only question is: how many of the $E$ values can we make odd in $p$?

When there is at least one odd-frequency value

If $O > 0$, then parity is flexible. An odd-frequency value can be split with either parity on $p$, and it still contributes exactly $1$ either way. So it can absorb the annoying parity issue.

That means we can make every even-frequency value good: give each one an odd count in $p$. Since each such value appears at least twice, giving it one copy is valid, and then extra copies can be added in pairs without changing parity. The odd-frequency values handle whatever leftover parity is needed.

So if $O > 0$, the answer is:

$O + 2E$.

When all frequencies are even

Now suppose $O = 0$. Every value has even total count.

If we make $m$ values good, then exactly those $m$ values have odd counts in $p$. All other values have even counts in $p$. Therefore the parity of $|p|$ is the same as the parity of $m$.

But $|p| = n$, so we must have:

$m$.

We want the largest possible $m <= E$ with that parity:

• if $E$ has the same parity as $n$, use $m = E$;
• otherwise, use $m = E - 1$.

Then the answer is $2m$.

This is the whole trick. The problem looks like it wants a complicated construction, but the output is only the maximum value, so parity does basically all the work. Classic Codeforces: count stuff, stare at parity, win.

Algorithm

For each test case:

1. Count frequencies of all values.
2. Count how many nonzero frequencies are odd: $O$.
3. Count how many nonzero frequencies are even: $E$.
4. If $O > 0$, print $O + 2E$.
5. Otherwise, print $2E$ if $E$ and $n$ have the same parity, else print $2(E-1)$.

Complexity is $O(n)$ per test case, because values are between $1$ and $2n$.