Maximize the Earning

 Problem:-

Napoleon choosed a city for Advertising his company's product. There are $S$ streets in that city. Each day he travels one street. There are $N$ buildings in a street which are located at points $1,2,\mathrm{3....}N($$respectively)$. Each building has some height(Say $h$ meters). Napoleon stands at point $0$. His height is $0.5m$. Now Napoleon starts communicating with the people of each building. He can communicate with people of a particular building only if he can see that building. If he succeed to communicate with any particular building then his boss gives him $Rrupees$. i.e. if he communicates with $K$ buildings in a day, then he will earn $K\ast Rrupees$. Now Napoleon wants to know his maximum Earning for each day.

Note: All the points are on Strainght Line and Napoleon is always standing at point 0.

Input:
First line of input contains an integer $S$, denoting no of streets in the city.
Details for each street is described in next two lines.
First line contains two integers $N$ and $R$$,$ denoting no of buildings in the Street and earning on communicating with a building.
Second line contains $N$ integers, denoting height of  building.

Output:
Print $S$ Lines, each containing maximum earning in $i^{th}$ street.
    
Constraints:
    $1\le N\le {10}^4$
    $1\le h\le {10}^9$
    $1\le S\le 100$
    $1\le R\le {10}^4$

Sample Input

2
6 3
8 2 3 11 11 10
5 12
12 20 39 45 89

Sample Output

6
60

Time Limit: 1

Memory Limit: 256

Source Limit:

Explanation

There are two streets in the city.

The first street has $6$ buildings and the earning of Napoleon for communicating with each building is $3rupees$.

Height of buildings are $8,2,3,11,11,10$ respectively.

As Chef is standing at point $0$, he will be able to see only 1st and 4th building.

So his total earning will be $3\ast 2=6rupees$ in that street

Similarly for 2nd street his earning will be $60rupees$

Code:-

#include<stdio.h>

int main()

{

    long int n,s,r,i;

    long long int max,h,building;

    scanf("%ld",&s);

    

    while(s--)

    {

        scanf("%ld%ld",&n,&r);

         max=0;

        building=0;

    for(i=0;i<n;i++)

        {

      scanf("%lld",&h);

      if(h>max)

           {

               max=h;

               building++;

           }

        }

        printf("%lld\n",r*building);

    }

    return 0;

}