<div class="_1l1MA">We can solve the problem using binary lifting.</div>

<div class="_1l1MA">For each player with id <code>x</code> and for every <code>i</code> in the range <code>[0, ceil(log₂k)]</code>, we can determine the last receiver's id and compute the sum of player ids who receive the ball after <code>2ⁱ</code> passes, starting from <code>x</code>.</div>

<div class="_1l1MA">Let <code>last_receiver[x][i] =</code> the last receiver's id after <code>2ⁱ</code> passes, and <code>sum[x][i] =</code> the sum of player ids who receive the ball after <code>2ⁱ</code> passes. For all <code>x</code> in the range <code>[0, n - 1]</code>, <code>last_receiver[x][0] = receiver[x]</code>, and <code>sum[x][0] = receiver[x]</code>.</div>

<div class="_1l1MA">Then for <code>i</code> in range <code>[1, ceil(log₂k)]</code>, <code>last_receiver[x][i] = last_receiver[last_receiver[x][i - 1]][i - 1]</code> and <code>sum[x][i] = sum[x][i - 1] + sum[last_receiver[x][i - 1]][i - 1]</code>, for all <code>x</code> in the range <code>[0, n - 1]</code>.</div>

<div class="_1l1MA">Starting from each player id <code>x</code>, we can now go through the powers of <code>2</code> in the binary representation of <code>k</code> and make jumps corresponding to each power, using the pre-computed values, to compute <code>f(x)</code>.</div>

<div class="_1l1MA">The answer is the maximum <code>f(x)</code> from each player id.</div>