/* Sample solution for NCPC'06: Jezzball
 * Author: Per Austrin
 */
#include <algorithm>
#include <cstdio>
#include <cmath>

using namespace std;

const int ray_speed = 200;
const int width = 1024;
const int height = 768;
const int timeout = 10000;
const double eps = 1e-6;

double find_teh_secret_time_of_fire(int rx, int ry, int W, int H,
				    int n, int *x, int *y, int *vx, int *vy) {
  double t = 0;
  double ext_time = max(1.0*rx / ray_speed, 1.0*(W-rx) / ray_speed);
  int fine = 0, pos = 0;
  while (t <= timeout && fine < n) {
    for (double curt = t; curt < t + ext_time && t <= timeout; ) {
      int wx = vx[pos], wy = vy[pos];
      double cx = fmod(x[pos] + wx * curt, 2*W);
      double cy = fmod(y[pos] + wy * curt, 2*H);
      if (cx < 0) cx = -cx, wx = -wx;
      if (cy < 0) cy = -cy, wy = -wy;
      if (cx > W) cx = 2*W - cx, wx = -wx;
      if (cy > H) cy = 2*H - cy, wy = -wy;      
      double dt = (ry - cy) / wy;
      if (fabs(dt) < eps) {
	if ((curt - t)*ray_speed > fabs(cx - rx) + eps)
	  curt = t = curt - fabs(cx - rx) / ray_speed, fine = 0;
	else
	  curt += 2*eps;
      } else if (dt < 0) {
	if (wy > 0) curt += (H-cy) / wy + 2*eps;
	else curt += -cy/wy + 2*eps;
      } else
	curt += dt;   
    }
    ++fine;
    ++pos %= n;
  }
  return t;
}

int main(void) {
  for (int n; scanf("%d", &n) && n; ) {
    int rx, ry;
    int x[30], y[30], vx[30], vy[30];
    scanf("%d%d", &rx, &ry);
    for (int i = 0; i < n; ++i)
      scanf("%d%d%d%d", x+i, y+i, vx+i, vy+i);
    double ans = min(find_teh_secret_time_of_fire(rx, ry, width, height,
						  n, x, y, vx, vy),
		     find_teh_secret_time_of_fire(ry, rx, height, width,
						  n, y, x, vy, vx));
    if (ans > timeout) printf("Never\n");
    else printf("%.5lf\n", ans);
  }
  return 0;
}