Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int dblcmp(double x) {
  if (abs(x) < 1e-8) return 0;
  return (x > 0) ? 1 : -1;
}
inline double sqr(double x) { return x * x; }
const double pi = acos(-1.0);
struct point {
  double x, y;
  point() {}
  point(double _x, double _y) : x(_x), y(_y){};
  void input() { scanf("%lf%lf", &x, &y); }
  void output() { printf("%.2f %.2f\n", x, y); }
  bool operator==(point a) const {
    return dblcmp(a.x - x) == 0 && dblcmp(a.y - y) == 0;
  }
  bool operator<(point a) const {
    return dblcmp(a.x - x) == 0 ? dblcmp(y - a.y) < 0 : x < a.x;
  }
  double len() { return hypot(x, y); }
  double len2() { return x * x + y * y; }
  double distance(point p) { return hypot(x - p.x, y - p.y); }
  point add(point p) { return point(x + p.x, y + p.y); }
  point sub(point p) { return point(x - p.x, y - p.y); }
  point mul(double b) { return point(x * b, y * b); }
  point div(double b) { return point(x / b, y / b); }
  double dot(point p) { return x * p.x + y * p.y; }
  double det(point p) { return x * p.y - y * p.x; }
  double rad(point a, point b) {
    point p = *this;
    return fabs(atan2(fabs(a.sub(p).det(b.sub(p))), a.sub(p).dot(b.sub(p))));
  }
  point trunc(double r) {
    double l = len();
    if (!dblcmp(l)) return *this;
    r /= l;
    return point(x * r, y * r);
  }
  point rotleft() { return point(-y, x); }
  point rotright() { return point(y, -x); }
  point rotate(point p, double angle) {
    point v = this->sub(p);
    double c = cos(angle), s = sin(angle);
    return point(p.x + v.x * c - v.y * s, p.y + v.x * s + v.y * c);
  }
};
struct line {
  point a, b;
  line() {}
  line(point _a, point _b) {
    a = _a;
    b = _b;
  }
  bool operator==(line v) { return (a == v.a) && (b == v.b); }
  line(point p, double angle) {
    a = p;
    if (dblcmp(angle - pi / 2) == 0) {
      b = a.add(point(0, 1));
    } else {
      b = a.add(point(1, tan(angle)));
    }
  }
  line(double _a, double _b, double _c) {
    if (dblcmp(_a) == 0) {
      a = point(0, -_c / _b);
      b = point(1, -_c / _b);
    } else if (dblcmp(_b) == 0) {
      a = point(-_c / _a, 0);
      b = point(-_c / _a, 1);
    } else {
      a = point(0, -_c / _b);
      b = point(1, (-_c - _a) / _b);
    }
  }
  void input() {
    a.input();
    b.input();
  }
  void adjust() {
    if (b < a) swap(a, b);
  }
  double length() { return a.distance(b); }
  double angle() {
    double k = atan2(b.y - a.y, b.x - a.x);
    if (dblcmp(k) < 0) k += pi;
    if (dblcmp(k - pi) == 0) k -= pi;
    return k;
  }
  int relation(point p) {
    int c = dblcmp(p.sub(a).det(b.sub(a)));
    if (c < 0) return 1;
    if (c > 0) return 2;
    return 3;
  }
  bool pointonseg(point p) {
    return dblcmp(p.sub(a).det(b.sub(a))) == 0 &&
           dblcmp(p.sub(a).dot(p.sub(b))) <= 0;
  }
  bool parallel(line v) { return dblcmp(b.sub(a).det(v.b.sub(v.a))) == 0; }
  int segcrossseg(line v) {
    int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));
    int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));
    int d3 = dblcmp(v.b.sub(v.a).det(a.sub(v.a)));
    int d4 = dblcmp(v.b.sub(v.a).det(b.sub(v.a)));
    if ((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2;
    return (d1 == 0 && dblcmp(v.a.sub(a).dot(v.a.sub(b))) <= 0 ||
            d2 == 0 && dblcmp(v.b.sub(a).dot(v.b.sub(b))) <= 0 ||
            d3 == 0 && dblcmp(a.sub(v.a).dot(a.sub(v.b))) <= 0 ||
            d4 == 0 && dblcmp(b.sub(v.a).dot(b.sub(v.b))) <= 0);
  }
  int linecrossseg(line v) {
    int d1 = dblcmp(b.sub(a).det(v.a.sub(a)));
    int d2 = dblcmp(b.sub(a).det(v.b.sub(a)));
    if ((d1 ^ d2) == -2) return 2;
    return (d1 == 0 || d2 == 0);
  }
  int linecrossline(line v) {
    if ((*this).parallel(v)) {
      return v.relation(a) == 3;
    }
    return 2;
  }
  point crosspoint(line v) {
    double a1 = v.b.sub(v.a).det(a.sub(v.a));
    double a2 = v.b.sub(v.a).det(b.sub(v.a));
    return point((a.x * a2 - b.x * a1) / (a2 - a1),
                 (a.y * a2 - b.y * a1) / (a2 - a1));
  }
  double dispointtoline(point p) {
    return fabs(p.sub(a).det(b.sub(a))) / length();
  }
  double dispointtoseg(point p) {
    if (dblcmp(p.sub(b).dot(a.sub(b))) < 0 ||
        dblcmp(p.sub(a).dot(b.sub(a))) < 0) {
      return min(p.distance(a), p.distance(b));
    }
    return dispointtoline(p);
  }
  point lineprog(point p) {
    return a.add(b.sub(a).mul(b.sub(a).dot(p.sub(a)) / b.sub(a).len2()));
  }
  point symmetrypoint(point p) {
    point q = lineprog(p);
    return point(2 * q.x - p.x, 2 * q.y - p.y);
  }
};
int N;
line bas[110];
point ret[110];
int main() {
  scanf("%d", &N);
  if (N < 5) {
    puts("No solution");
    return 0;
  }
  double minOO = 249.66 / tan(acos(-1.0) / N);
  double maxOO = 250.33 / tan(acos(-1.0) / N);
  int i;
  srand(23333);
  for (i = 1; i <= N; i++) {
    int factor = rand();
    double len = (maxOO * factor + minOO * (32768 - factor)) / 32768;
    bas[i] = line(point(0, 0),
                  point(len * cos(2 * pi / N * i), len * sin(2 * pi / N * i)));
    bas[i].a = bas[i].a.rotate(bas[i].b, pi / 2);
  }
  for (i = 1; i <= N; i++) {
    ret[i] = bas[i].crosspoint(bas[i % N + 1]);
    printf("%.6f %.6f\n", ret[i].x, ret[i].y);
  }
}