【code force】模拟 746C

The tram in Berland goes along a straight line from the point 0 to the point s and back, passing 1 meter per t₁ seconds in both directions. It means that the tram is always in the state of uniform rectilinear motion, instantly turning around at points x = 0 and x = s.

Igor is at the point x₁. He should reach the point x₂. Igor passes 1meter per t₂ seconds.

Your task is to determine the minimum time Igor needs to get from the point x₁ to the point x₂, if it is known where the tram is and in what direction it goes at the moment Igor comes to the point x₁.

Igor can enter the tram unlimited number of times at any moment when his and the tram's positions coincide. It is not obligatory that points in which Igor enter and exit the tram are integers. Assume that any boarding and unboarding happens instantly. Igor can move arbitrary along the line (but not faster than 1 meter per t₂seconds). He can also stand at some point for some time.

Input

The first line contains three integers s, x₁ and x₂ (2 ≤ s ≤ 1000, 0 ≤ x₁, x₂ ≤ s, x₁ ≠ x₂) — the maximum coordinate of the point to which the tram goes, the point Igor is at, and the point he should come to.

The second line contains two integers t₁ and t₂ (1 ≤ t₁, t₂ ≤ 1000) — the time in seconds in which the tram passes 1 meter and the time in seconds in which Igor passes 1 meter.

The third line contains two integers p and d (1 ≤ p ≤ s - 1, d is either1 or ) — the position of the tram in the moment Igor came to the point x₁ and the direction of the tram at this moment. If , the tram goes in the direction from the point s to the point 0. If d = 1, the tram goes in the direction from the point 0 to the point s.

Output

Print the minimum time in seconds which Igor needs to get from the point x₁ to the point x₂.

Example

Input

4 2 43 41 1

Output

8

Input

5 4 01 23 1

Output

7

Note

In the first example it is profitable for Igor to go by foot and not to wait the tram. Thus, he has to pass 2 meters and it takes 8seconds in total, because he passes 1 meter per 4 seconds.

In the second example Igor can, for example, go towards the point x₂and get to the point 1 in 6 seconds (because he has to pass 3 meters, but he passes 1 meters per 2 seconds). At that moment the tram will be at the point 1, so Igor can enter the tram and pass 1 meter in 1second. Thus, Igor will reach the point x₂ in 7 seconds in total.

题意：一条直线上坐标从0到s，某人从x1到x2;向前走一米需要t2秒,有一辆车子起始位置为p,
向前走一米需要t1秒，车子方向d==1时为正，d==-1时为负，人可以原地不动，等与车子相遇时，人可以坐上车子，问人从x1到x2最少需要多少时间

code：

#include<cstdio>#include<algorithm>using namespace std;int main(){int s,x1,x2,lu1,lu2;int p,d,t1,t2;while(~scanf("%d%d%d",&s,&x1,&x2)){scanf("%d%d",&t1,&t2);scanf("%d%d",&p,&d);if(x2>x1){    lu1=(x2-x1)*t2;//人走所需时间if(d==1){//正向 if(p>x1) lu2=(s-p+s+x2)*t1;else lu2=(x2-p)*t1;}else if(d==-1) lu2=(p+x2)*t1;        if(lu1>lu2)printf("%d\n",lu2);else printf("%d\n",lu1); }else{lu1=(x1-x2)*t2;if(d==1) lu2=(s-p+s-x2)*t1;else if(d==-1){if(p<x1)lu2=(p+s+s-x2)*t1;else lu2=(p-x2)*t1;}if(lu1>lu2)printf("%d\n",lu2);else printf("%d\n",lu1); }}return 0; }