Eccoci.

This problem is based on (but not identical to) the scoring for the card game Cribbage.

Consider a normal pack of $52$ cards. A Hand is a selection of one or more of these cards.

For each Hand the Hand score is the sum of the values of the cards in the Hand where the value of Aces is $1$ and the value of court cards (Jack, Queen, King) is $10$.

The Cribbage score is obtained for a Hand by adding together the scores for:

• Pairs. A pair is two cards of the same rank. Every pair is worth $2$ points.
• Runs. A run is a set of at least $3$ cards whose ranks are consecutive, e.g. 9, 10, Jack. Note that Ace is never high, so Queen, King, Ace is not a valid run. The number of points for each run is the size of the run. All locally maximum runs are counted. For example, 2, 3, 4, 5, 7, 8, 9 the two runs of 2, 3, 4, 5 and 7, 8, 9 are counted but not 2, 3, 4 or 3, 4, 5.
• Fifteens. A fifteen is a combination of cards that has value adding to $15$. Every fifteen is worth $2$ points. For this purpose the value of the cards is the same as in the Hand Score.

For example, $(5 \spadesuit, 5 \clubsuit, 5 \diamondsuit, K \heartsuit)$ has a Cribbage score of $14$ as there are four ways that fifteen can be made and also three pairs can be made.

The example $( A \diamondsuit, A \heartsuit, 2 \clubsuit, 3 \heartsuit, 4 \clubsuit, 5 \spadesuit)$ has a Cribbage score of $16$: two runs of five worth $10$ points, two ways of getting fifteen worth $4$ points and one pair worth $2$ points. In this example the Hand score is equal to the Cribbage score.

Find the number of Hands in a normal pack of cards where the Hand score is equal to the Cribbage score.