In this problem we consider a very simplified model of Barcelona city.

Barcelona can be represented as a plane with streets of kind x = c and y = c for every integer c (that is, the rectangular grid). However, there is a detail which makes Barcelona different from Manhattan. There is an avenue called Avinguda Diagonal which can be represented as a the set of points (x, y) for which ax + by + c = 0.

One can walk along streets, including the avenue. You are given two integer points A and B somewhere in Barcelona. Find the minimal possible distance one needs to travel to get to B from A.

Input

The first line contains three integers a, b and c (-10^9≤ a, b, c≤ 10^9, at least one of a and b is not zero) representing the Diagonal Avenue.

The next line contains four integers x_1, y_1, x_2 and y_2 (-10^9≤ x_1, y_1, x_2, y_2≤ 10^9) denoting the points A = (x_1, y_1) and B = (x_2, y_2).

Output

Find the minimum possible travel distance between A and B. Your answer is considered correct if its absolute or relative error does not exceed 10^{-6}.

Formally, let your answer be a, and the jury's answer be b. Your answer is accepted if and only if \frac{|a - b|}{max{(1, |b|)}} ≤ 10^{-6}.

Examples

Input


1 1 -3
0 3 3 0


Output


4.2426406871


Input


3 1 -9
0 3 3 -1


Output


6.1622776602

Note

The first example is shown on the left picture while the second example us shown on the right picture below. The avenue is shown with blue, the origin is shown with the black dot.

<image>
#include <bits/stdc++.h>
using namespace std;
typedef struct {
  double x, y;
} point;
double dis(point a, point b) {
  return (sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)));
}
int main() {
  double a, b, c;
  cin >> a >> b >> c;
  point A, B;
  cin >> A.x >> A.y >> B.x >> B.y;
  if (b == 0) {
    point A1, B1;
    A1.x = -(double)c / a;
    A1.y = A.y;
    B1.x = -(double)c / a;
    B1.y = B.y;
    double mandis = abs(A.x - B.x) + abs(A.y - B.y);
    double AA1, BB1, A1B1;
    AA1 = dis(A, A1);
    BB1 = dis(B, B1);
    A1B1 = dis(A1, B1);
    double ans = min(mandis, (AA1 + BB1 + A1B1));
    printf("%.10lf", ans);
  } else if (a == 0) {
    point A1, B1;
    A1.x = A.x;
    A1.y = -(double)c / b;
    B1.y = -(double)c / b;
    B1.x = B.x;
    double mandis = abs(A.x - B.x) + abs(A.y - B.y);
    double AA1, BB1, A1B1;
    AA1 = dis(A, A1);
    BB1 = dis(B, B1);
    A1B1 = dis(A1, B1);
    double ans = min(mandis, (AA1 + BB1 + A1B1));
    printf("%.10lf", ans);
  } else {
    point A1, A2, B1, B2;
    A1.x = A.x;
    A1.y = -((double)(a * A.x + c) / b);
    A2.x = -((double)(b * A.y + c) / a);
    A2.y = A.y;
    B1.x = B.x;
    B1.y = -((double)(a * B.x + c) / b);
    B2.x = -((double)(b * B.y + c) / a);
    B2.y = B.y;
    double mandis = abs(A.x - B.x) + abs(A.y - B.y);
    double AA1, AA2, BB1, BB2, A1B1, A1B2, A2B1, A2B2;
    AA1 = dis(A, A1);
    AA2 = dis(A, A2);
    BB1 = dis(B, B1);
    BB2 = dis(B, B2);
    A1B1 = dis(A1, B1);
    A1B2 = dis(A1, B2);
    A2B1 = dis(A2, B1);
    A2B2 = dis(A2, B2);
    double path1, path2, path3, path4;
    path1 = AA1 + A1B1 + BB1;
    path2 = AA1 + A1B2 + BB2;
    path3 = AA2 + A2B1 + BB1;
    path4 = AA2 + A2B2 + BB2;
    vector<double> paths;
    paths.push_back(path1);
    paths.push_back(path2);
    paths.push_back(path3);
    paths.push_back(path4);
    sort(paths.begin(), paths.end());
    double ans = min(paths[0], mandis);
    printf("%.10lf", ans);
  }
}
