18. Four different prime numbers $\mathrm{a}, \mathrm{~b}, \mathrm{c}, \mathrm{~d}$ satisfy the following properties:
(1) $a+b+c+d$ is also a prime number; (2) The sum of two of $\mathrm{a}, \mathrm{~b}, \mathrm{c}, \mathrm{~d}$ is also a prime number; (3) The sum of three of $\mathrm{a}, \mathrm{~b}, \mathrm{c}, \mathrm{~d}$ is also a prime number. The smallest value of $a+b+c+d$ that satisfies the conditions is $\qquad$ -
【Analysis】By parity, if the four numbers are all odd, then their sum must be even, which clearly cannot be a prime number. Therefore, among the four numbers, there must be a 2, let's say $a=2$.
Among them, the sum of some three numbers is also a prime number, so the sum of the other three odd numbers excluding 2 must be a prime number.
Thus, $b+c+d$ and $a+b+c+d$ are a pair of prime numbers with a difference of 2, i.e., twin primes. The smallest value of $b+c+d$ is $3+5+7=15$, which means we need to find the smallest twin primes greater than 15. The smallest pair is $17, 19$, but unfortunately, 17 cannot be decomposed into the sum of three different primes. The next pair is 29, 31, which can be achieved with the four numbers $2, 3, 7, 19$, where $2+3+7+19=31$, $3+7+19=29$, and $2+3=5$, meeting the requirements. Therefore, the minimum value of $a+b+c+d$ is 31.