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Let $n! = 1 \times 2 \times 3 \times \dots \times n$ for integer $n \geq 1$. If $p = 1! (2 \times 2!) + (3 \times 3!) + \dots + (10 \times 10!)$, then $p+2$ when divided by $11!$ leaves a remainder of

1. $10$
2. $0$
3. $7$
4. $1$

1