Given a natural number x and n different coin values c_{1} …
c_{n}, compute in how many ways it is possible to achieve change x by using each
value at most twice. Here, two coins with the same value are
considered different.

For example, if x = 4 and the available values are 1 and 2, then there are three ways to achieve it: 1 + 1′ + 2, 1 + 1′ + 2′, and also 2 + 2′.

Input

Input consists of several cases.
Every case begins with x and n,
followed by c_{1} … c_{n}.
Assume 1 ≤ n ≤ 20,
1 ≤ c_{i} ≤ x ≤ 1000,
and that all c_{i} are different.

Output

For every case print, in lexicographic order, all possible ways to exactly achieve change x by using each value at most twice. Print every solution with its values sorted from small to big. In doing that, assume 1 < 1′ < 2 < 2′ < …. Use “1p” to print 1′, etcetera. Print a line with 10 dashes at the end of every case.

Hint

A simply pruned backtracking should be enough.

Public test cases

**Input**

4 2 1 2 400 1 200 400 1 300 5 3 4 2 1 5 5 1 2 3 4 5

**Output**

4 = 1 + 1p + 2 4 = 1 + 1p + 2p 4 = 2 + 2p ---------- 400 = 200 + 200p ---------- ---------- 5 = 1 + 2 + 2p 5 = 1 + 4 5 = 1 + 4p 5 = 1p + 2 + 2p 5 = 1p + 4 5 = 1p + 4p ---------- 5 = 1 + 1p + 3 5 = 1 + 1p + 3p 5 = 1 + 2 + 2p 5 = 1 + 4 5 = 1 + 4p 5 = 1p + 2 + 2p 5 = 1p + 4 5 = 1p + 4p 5 = 2 + 3 5 = 2 + 3p 5 = 2p + 3 5 = 2p + 3p 5 = 5 5 = 5p ----------

Information

- Author
- Salvador Roura
- Language
- English
- Translator
- Albert Atserias
- Original language
- Catalan
- Other languages
- Catalan
- Official solutions
- C++
- User solutions
- C++