Covered Path

Description

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 v₂meters 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

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

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 = 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.

描述：

一個小車起始速度為v1，末速度為v2,給出時間t，和最大加速度（-d--d）

求小車在 t 時間內的最大路程。

正確解法：

時間從 t==2 開始列舉，從d— -d開始列舉加速度。

當滿足 現在的速度v 加上加速度 再減去剩餘時間*d <=v2 時，此時的加速度就是目前所能加的最大加速度。

萬一小車加速，要保證它在剩餘時間內可以減速到 v2

（第一次見到列舉加速度，要好好記下來）

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<string>
 6 #include<cstring>
 7 using namespace std;    
 8 int b[1000010] = {0};
 9 int main()
10 {
11     int v1, v2,t,d,v;
12     cin >> v1 >> v2 >> t >> d;
13     long long ans = v1;
14     v = v1;    t=t-1;
15     while (t) {
16         for (int i = d; i >= -d; i--)
17             if (v + i - (t - 1)*d <= v2)
18             {
19                 ans += v + i;
20                 t--;
21                 v = v + i;
22                 break;
23             }
24     }
25     cout << ans << endl;
26     return 0;
27 }