Polycarpus loves convex polygons, especially if all their angles are the same and all their sides are different. Draw for him any such polygon with the given number of vertexes.

Input

The input contains a single integer n (3 ≤ n ≤ 100) — the number of the polygon vertexes.

Output

Print n lines, containing the coordinates of the vertexes of the n-gon "xi yi" in the counter clockwise order. The coordinates of the vertexes shouldn't exceed 106 in their absolute value. The side lengths should fit within limits [1, 1000] (not necessarily integer). Mutual comparing sides and angles of your polygon during the test will go with precision of 10 - 3.

If there is no solution, print "No solution" (without the quotes).

Examples

Input

8


Output

1.000 0.000
7.000 0.000
9.000 2.000
9.000 3.000
5.000 7.000
3.000 7.000
0.000 4.000
0.000 1.000
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1000;
const double eps = 1e-8, inf = 1e50, Pi = acos(-1.0);
struct point {
  double x, y;
  point() {}
  point(double _x, double _y) {
    x = _x;
    y = _y;
  }
  point operator-(const point &ne) const { return point(x - ne.x, y - ne.y); }
  point operator+(const point ne) const { return point(x + ne.x, y + ne.y); }
  point operator*(const double t) const { return point(x * t, y * t); }
  point operator/(const double t) const {
    if (((t) > eps ? 1 : ((t) < -eps ? (-1) : (0))) == 0) exit(1);
    return point(x / t, y / t);
  }
};
struct line {
  point a, b;
  line() {}
  line(point _a, point _b) {
    a = _a;
    b = _b;
  }
};
double dmult(point a, point b) { return a.x * b.x + a.y * b.y; }
double lenth(point a) { return sqrt(dmult(a, a)); }
inline double xmult(point o, point a, point b) {
  return (a.x - o.x) * (b.y - o.y) - (b.x - o.x) * (a.y - o.y);
}
inline double xmult(double x1, double y1, double x2, double y2) {
  return x1 * y2 - x2 * y1;
}
inline double dmult(point o, point a, point b) {
  return (a.x - o.x) * (b.x - o.x) + (a.y - o.y) * (b.y - o.y);
}
inline double xmult(point a, point b) { return a.x * b.y - a.y * b.x; }
point line_intersection(line u, line v) {
  point ret = u.a;
  double t = xmult(u.a - v.a, v.b - v.a) / xmult(u.b - u.a, v.b - v.a);
  t = fabs(t);
  ret.x += (u.b.x - u.a.x) * t;
  ret.y += (u.b.y - u.a.y) * t;
  return ret;
}
point line_intersection2(line u, line v) {
  double a1 = u.b.y - u.a.y, b1 = u.a.x - u.b.x;
  double c1 = u.b.y * (-b1) - u.b.x * a1;
  double a2 = v.b.y - v.a.y, b2 = v.a.x - v.b.x;
  double c2 = v.b.y * (-b2) - v.b.x * a2;
  double D = xmult(a1, b1, a2, b2);
  return point(xmult(b1, c1, b2, c2) / D, xmult(c1, a1, c2, a2) / D);
}
void counter(double &x, double &y, double th) {
  double tx = x, ty = y;
  x = tx * cos(th) - ty * sin(th);
  y = tx * sin(th) + ty * cos(th);
  return;
}
int main() {
  int n;
  scanf("%d", &n);
  if (n <= 4) {
    printf("No solution\n");
    return 0;
  }
  point tmp, way(-800.0, 0), po[105];
  double delta = 0.999995, th = -2.0 * Pi / n;
  po[0] = point(0, 0);
  for (int i = 1; i <= n - 2; i++) {
    counter(way.x, way.y, th);
    way = way * delta;
    po[i] = po[i - 1] + way;
  }
  counter(way.x, way.y, th);
  way = way * delta;
  tmp = po[n - 2] + way;
  po[n - 1] = line_intersection(line(po[n - 2], tmp),
                                line(point(-1.0, 0.0), point(1.0, 0.0)));
  po[n] = po[0];
  for (int i = n - 1; i >= 0; i--) {
    printf("%lf %lf\n", po[i].x, po[i].y);
  }
  return 0;
}
