* Antonio foloseste mult expresia "pe mangleala" si cuvantul "gen".

* **Subtask 1 (30 puncte)**: $1 ≤ N ≤ 20$, $M = 0$
* **Subtask 2 (20 puncte)**: <tex> 1 \leq N \cdot M \leq 4.000.000$ </tex>, $1 &le; M &le; 4000$ si nu vor exista $2$ angajati cu acelasi sef direct
* **Subtask 4 (50 puncte)**: $1 &le; N &le; 100000$

* **Subtask 1 (30 puncte)**: <tex> 1 \leq N \leq 20 </tex>, <tex> M = 0 </tex>
* **Subtask 2 (20 puncte)**: <tex> 1 \leq N \cdot M \leq 4.000.000$ </tex>, <tex> 1 \leq M \leq 4.000 </tex> si nu vor exista $2$ angajati cu acelasi sef direct
* **Subtask 4 (50 puncte)**: <tex> 1 \leq N \leq 100.000 </tex>

h2. Exemplu

Exista $5$ posibilitati de a concedia angajatii (considerand si ordinea concedierilor):

# Angajatul **$1$** cu probabilitatea <tex> \(\frac{1}{3}\) </tex>, in acest caz Zetul va bea <tex> val_1 </tex> shot-uri;
# Angajatii **$2$**, **$1$** cu probabilitatea <tex> \(\frac{1}{3} \cdot \frac{1}{2}\) </tex>, in acest caz Zetul va bea <tex> val_2 + val_1 </tex> shot-uri;
# Angajatii **$3$**, **$1$** cu probabilitatea <tex> \(\frac{1}{3} \cdot \frac{1}{2}\) </tex>, in acest caz Zetul va bea <tex> val_3 + val_1 </tex> shot-uri;
# Angajatii **$2$**, **$3$**, **$1$** cu probabilitatea <tex> \(\frac{1}{3} \cdot \frac{1}{2} \cdot 1\) </tex>, in acest caz Zetul va bea <tex> val_2 + val_3 + val_1 </tex> shot-uri;
# Angajatii **$3$**, **$2$**, **$1$** cu probabilitatea <tex> \(\frac{1}{3} \cdot \frac{1}{2} \cdot 1\) </tex>, in acest caz Zetul va bea <tex> val_3 + val_2 + val_1 </tex> shot-uri;

# Angajatul **$1$** cu probabilitatea <tex> \(\frac{1}{3}\) </tex> - in acest caz Zetul va bea <tex> val_1 </tex> shot-uri;
# Angajatii **$2$**, **$1$** cu probabilitatea <tex> \(\frac{1}{3} \cdot \frac{1}{2}\) </tex> - in acest caz Zetul va bea <tex> val_2 + val_1 </tex> shot-uri;
# Angajatii **$3$**, **$1$** cu probabilitatea <tex> \(\frac{1}{3} \cdot \frac{1}{2}\) </tex> - in acest caz Zetul va bea <tex> val_3 + val_1 </tex> shot-uri;
# Angajatii **$2$**, **$3$**, **$1$** cu probabilitatea <tex> \(\frac{1}{3} \cdot \frac{1}{2} \cdot 1\) </tex> - in acest caz Zetul va bea <tex> val_2 + val_3 + val_1 </tex> shot-uri;
# Angajatii **$3$**, **$2$**, **$1$** cu probabilitatea <tex> \(\frac{1}{3} \cdot \frac{1}{2} \cdot 1\) </tex> - in acest caz Zetul va bea <tex> val_3 + val_2 + val_1 </tex> shot-uri;

Asadar: