Covered Path

The on-board computer on Polycarp's car measured that the car speed at the beginning of some section of the path equals v1 meters per second, and in the end it is v2 meters per second. We know that this section of the route took exactly t seconds to pass.

Assuming that at each of the seconds the speed is constant, and between seconds the speed can change at most by d meters per second in absolute value (i.e., the difference in the speed of any two adjacent seconds does not exceed d in absolute value), find the maximum possible length of the path section in meters.

Input

The first line contains two integers v1 and v2 (1 ≤ v1, v2 ≤ 100) — the speeds in meters per second at the beginning of the segment and at the end of the segment, respectively.

The second line contains two integers t (2 ≤ t ≤ 100) — the time when the car moves along the segment in seconds, d (0 ≤ d ≤ 10) — the maximum value of the speed change between adjacent seconds.

It is guaranteed that there is a way to complete the segment so that:

• the speed in the first second equals v1,
• the speed in the last second equals v2,
• the absolute value of difference of speeds between any two adjacent seconds doesn't exceed d.

Output

Print the maximum possible length of the path segment in meters.

Sample test(s)

input

5 6
4 2

output

26

input

10 10
10 0

output

100

Note

In the first sample the sequence of speeds of Polycarpus' car can look as follows: 5, 7, 8, 6. Thus, the total path is 5 + 7 + 8 + 6 = 26meters.

In the second sample, as d = 0, the car covers the whole segment at constant speed v = 10. In t = 10 seconds it covers the distance of 100 meters.

#####################################editorial############################

in this problem we 1st calculate the maximum speed a driver can have at any instant of time..

here maximum speed means speed which can be possible to gain and fromk that speed we can reach destination at given spees .. and if our required speed is less than increase speed by amount inn the range 1 to d so that reach maximum speed .. 

similarly if speed is more than maximum speed at any instant than decrease speed but try  to  maximum   speed he can accheave at that instant ......

##################################### code ##################################

#include<iostream>
using namespace std;
#include<bits/stdc++.h>
#define first ff
#define second ss
typedef long long int lli;
typedef long int li;
int  main()
 
 {
 // freopen("abc.txt","r",stdin);
  
  int   v1,v2;
     cin>>v1>>v2;
     
    int t,d;
    cin>>t>>d;
   long long  int sum=v1;
   long long  int spd=v1;
    for(int i=2;i<t;i++)
     {
     long long  int max=v2+(t-i)*d;// maximu speed which can be possible at the it second 
      
      if(spd<max)
       {
          if(spd+d<=max) spd+=d;
          else spd=max;
          sum+=spd;
          
        
       }
       else
        {
           spd=max;
           sum+=max;
        }
          
           
         // cout<<i<<" sum "<<sum<<endl;
     }
      cout<<sum+v2<<endl;
    
  return  0;
   
 }