Question

solve the following cp problem.



Chef may buy a number of tables, which are large enough for any number of guests, but

the people sitting at each table must have consecutive numbers ― for any two

guests ii and j (i<j) sitting at the same table, guests i+1,i+2,...,j−1 must also sit at that

table. Chef would have liked to seat all guests at a single table; however, he noticed that

two guests i and j are likely to get into an argument if Fi=Fj and they are sitting at the

same table.

Each table costs K rupees. Chef defines the inefficiency of a seating arrangement as the

total cost of tables plus the number of guests who are likely to get into an argument with

another guest. Tell Chef the minimum possible inefficiency which he can achieve.

Input

• The first line of the input contains a single integer T denoting the number of test

cases. The description of T test cases follows.

• The first line of each test case contains two space-separated integers N and K.

• The second line contains N space-separated integers F1,F2,...,FN.

Output

For each test case, print a single line containing one integer ― the smallest possible

inefficiency.

Constraints

• 1≤T≤100

• 1≤N≤1,000

• 1≤K≤1,000

• 1≤Fi≤100 for each valid

• The sum of N across test cases is ≤5,000

Example Input

3

5 1

5 1 3 3 3

5 14

1 4 2 4 4

5 2

3 5 4 5 1

Example Output

3

17

4

Explanation

Example case 1: The optimal solution is to use three tables with groups of

guests [1,2,3], [4] and [5. None of the tables have any guests that are likely to get into

an argument, so the inefficiency is 3⋅K=3


Example case 2: The optimal solution is to seat all guests at one table. Then,

guests 2, 4 and 5 are likely to get into an argument with each other, so the inefficiency

is K+3=17​

Answer 1

Answer:

can't understand your question please send only one question please

Answer 2

can we combine both to  produce optimal

.