Infinite Work

Let $N=k+1$.

The infinite tree is mostly psychological warfare. The actual problem is finite: choose a boundary of cuts near the root, and decide on which day each cut happens.

What One Firing Does

Suppose node $v$ is directly fired on day $d$.

After that firing, the whole subtree of $v$ is gone. On later days, hiring grows it back from the top, one level per day. So by the end:

• if $d=n$, it contributes $0$ nodes;
• if $d=n-1$, it contributes $1$ node;
• if $d=n-2$, it contributes $3$ nodes;
• in general, it contributes $2^{n-d}-1$ nodes.

Write $m=n-d$. Then a fired node contributes $2^m-1$, where $0 <= m < n$.

Also, once a subtree has been fired, you cannot later cut inside it. Those students were already fired indirectly. When they come back, firing them again is illegal. No double-dipping, no funny business.

The Fired Nodes Are A Boundary

To end with exactly finite $k$, every infinite downward path must be cut somewhere. Otherwise infinitely many workers survive on that path's branch, which is very much not $k$.

The directly fired nodes therefore form an antichain covering all infinite paths. Equivalently, they are the leaves of a finite full binary tree rooted at student $1$.

If there are $L$ direct firings, this full binary tree has:

• $L$ leaves, the directly fired nodes;
• $L-1$ internal nodes, the nodes above the cuts that never get fired.

For fired leaf $i$, let $m_i=n-d_i$. Then the final number of active students is

$k=(L-1)+\sum_i(2^{m_i}-1)$.

The annoying terms cancel beautifully:

$k+1=\sum_i 2^{m_i}$.

So each direct firing is just one power of two in a representation of $k+1$, and the largest allowed power is $2^{n-1}$.

Minimizing Direct Firings

We need the fewest powers of two from

$1,2,4,\dots,2^{n-1}$

that sum to $N=k+1$.

Greedy is optimal: use the largest allowed power as much as possible, then use the binary representation of the remainder.

Let:

• $b=min(n-1, floor(log_2 N))$;
• $p=2^b$;
• $q=floor(N/p)$;
• $r=N mod p$.

The unique minimal multiset is:

• $q$ copies of $p$;
• one copy of $2^j$ for each set bit $j$ in $r$.

Thus

$L=q+popcount(r)$.

Counting

Now count strategies with this minimal $L$.

First choose the boundary shape. A full ordered binary tree with $L$ leaves is counted by

$Catalan_{L-1}$.

Second assign the chosen exponents to the $L$ leaves. The $q$ largest powers are identical. All smaller powers are distinct because they come from binary set bits. So the number of assignments is

$L! / q!$.

Therefore the answer is

$Catalan_{L-1} * L! / q! \pmod{10^9+7}$.

Use

$Catalan_m = C(2m,m)/(m+1)$.

Since $k <= 2 * 10^5$, factorials up to about $4 * 10^5$ are enough. The infinite tree was just a very dramatic way to ask for Catalan numbers and binary representation. Classic Codeforces nonsense, honestly.