Description

A traditional game is played between two players on a pool of n numbers (not necessarily distinguishing ones).

The first player will choose from the pool a number x1 lying in [a, b] (0 < a < b), which means a <= x1 <= b. Next the second player should choose a number y1 such that y1 – x1 lies in [a, b] (Attention! This implies y1 > x1 since a > 0). Then the first player should choose a number x2 such that x2 – y1 lies in [a, b]… The game ends when one of them cannot make a choice. Note that a player MUST NOT skip his turn.

A player’s score is determined by the numbers he has chose, by the way:

player1score = x1 + x2 + …
player2score = y1 + y2 + …

If you are player1, what is the maximum score difference (player1score – player2score) you can get? It is assumed that player2 plays perfectly.

Input

The first line contains a single integer t (1 <= t <= 20) indicating the number of test cases. Then follow the t cases. Each case contains exactly two lines. The first line contains three integers, n, a, b (2 <= n <= 10000, 0 < a < b <= 100); the second line contains n integers, the numbers in the pool, any of which lies in [-9999, 9999].
Output

For each case, print the maximum score difference player1 can get. Note that it can be a negative, which means player1 cannot win if player2 plays perfectly.
Sample Input

3
6 1 2
1 3 -2 5 -3 6
2 1 2
-2 -1
2 1 2
1 0

Sample Output

-3
0
1

题解

后面的人取的肯定要比前面的人取的要大，那么先排序，就可以依次取。

用f[m]表示如果先手取了第m个数，可以达到的最大分差。

那么最后一次取的分差就是本身。