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.

Examples

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 = 26 meters.

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.

#include<stdio.h>
int main()
{
    int v1,v2,v,t,d;
    int sum=0;
    int speed[110]; /*記錄每個時間段速度*/
    scanf("%d%d%d%d",&v1,&v2,&t,&d);
    speed[1]=v1;
    speed[t]=v2;
    for(int i=2;i<t;i++)
    {
        v=speed[i-1]+d;
        if(v>(t-i)*d+v2)
        v=(t-i)*d+v2;
        speed[i]=v;
    } 
    for(int i=1;i<=t;i++)
    sum+=speed[i];
    printf("%d",sum);
    return 0;
}

要得到通過的最長的路程，即要速度加到條件內的最大速度，並且在時間內減速回到v2。同時從v1，v2開始加速（兩邊往中間），如果從v1開始加的速度大於v2加的，那麼就不可以在時間內回到v2，最大的速度只能是從v2開始加的速度。
（國語と數學はあまりにも劣っていて、どのように説明するか分かりません）