Fix the value of <code>(subs[0] + subs[1]) % k</code> from the <code>k</code> possible values. Let it be <code>val</code>.

Let <code>dp[i]</code> store the maximum length of a subsequence with its last element <code>x</code> such that <code>x % k == i</code>.

Answer for a subsequence ending at index <code>y</code> is <code>dp[(k + val - (y % k)) % k] + 1</code>.