first for every I the I-th line has precisely one impediment — at hub (i,ai). You need to move a few hindrances so you can arrive at hub (n,106+1) from hub (1,0) by traveling through edges of this chart (you can't go through deterrents). Moving one deterrent to a nearby by edge free hub costs u or v coins, as underneath: In case there is a snag in the
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At first for every I the I-th line has precisely one impediment — at hub (i,ai). You need to move a few hindrances so you can arrive at hub (n,106+1) from hub (1,0) by traveling through edges of this chart (you can't go through deterrents). Moving one deterrent to a nearby by edge free hub costs u or v coins, as underneath:
In case there is a snag in the hub (i,j), you can utilize u coins to move it to (i−1,j) or (i+1,j), if such hub exists and in case there is no obstruction in that hub as of now.
In case there is a deterrent in the hub (i,j), you can utilize v coins to move it to (i,j−1) or (i,j+1), if such hub exists and in case there is no obstruction in that hub presently.
Note that you can't move hindrances outside the matrix. For instance, you can't move a hindrance from (1,1) to (0,1).
Allude to the image above for a superior agreement.
Presently you really wanted to work out the insignificant number of coins you wanted to spend to have the option to arrive at hub (n,106+1) from hub (1,0) by traveling through edges of this diagram without going through impediments.
Input
The primary line contains a solitary integer t (1≤t≤104) — the number of experiments.
The principal line of each experiment contains three integers n, u and v (2≤n≤100, 1≤u,v≤109) — the number of lines in the chart and the numbers of coins expected to move upward and on a level plane individually.
The second line of each experiment contains n integers a1,a2,… ,an (1≤ai≤106) — where
It's surefire that the amount of n over all experiments doesn't surpass 2⋅104.
Output
For each experiment, output a solitary integer — the negligible number of coins you wanted to spend to have the option to arrive at hub (n,106+1) from hub (1,0) by traveling through edges of this chart without going through impediments.
It tends to be shown that under the imperatives of the issue there is consistently a way of making such an outing conceivable
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