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;
const double PI = acos(double(-1));
const double INF = 1e10;
const double EPS = 1e-8;
inline double sqr(double x) { return x * x; }
struct PT {
  double x, y;
  PT() {}
  PT(double x, double y) : x(x), y(y) {}
  void in() { scanf("%lf%lf", &x, &y); }
};
bool operator<(const PT& p1, const PT& p2) {
  if (fabs(p1.x - p2.x) > EPS) return p1.x < p2.x;
  return p1.y + EPS < p2.y;
}
bool operator==(const PT& p1, const PT& p2) {
  return fabs(p1.x - p2.x) < EPS && fabs(p1.y - p2.y) < EPS;
}
PT operator+(PT p1, PT p2) { return PT(p1.x + p2.x, p1.y + p2.y); }
PT operator-(PT p1, PT p2) { return PT(p1.x - p2.x, p1.y - p2.y); }
PT operator*(PT p, double c) { return PT(p.x * c, p.y * c); }
PT operator/(PT p, double c) { return PT(p.x / c, p.y / c); }
double dis(PT p) { return sqrt(sqr(p.x) + sqr(p.y)); }
double dis2(PT p) { return sqr(p.x) + sqr(p.y); }
double dis2(PT p1, PT p2) { return sqr(p1.x - p2.x) + sqr(p1.y - p2.y); }
double dis(PT p1, PT p2) { return sqrt(dis2(p1, p2)); }
double vect(PT p1, PT p2) { return p1.x * p2.y - p2.x * p1.y; }
double scal(PT p1, PT p2) { return p1.x * p2.x + p1.y * p2.y; }
double getAngle(PT p) { return atan2(p.y, p.x); }
double vect(PT p, PT p1, PT p2) { return vect(p1 - p, p2 - p); }
double scal(PT p, PT p1, PT p2) { return scal(p1 - p, p2 - p); }
double getAngle(PT p, PT p1, PT p2) {
  return getAngle(p2 - p) - getAngle(p1 - p);
}
double getAngle180(PT p, PT p1, PT p2) {
  return acos(scal(p, p1, p2) / dis(p, p1) / dis(p, p2));
}
double disToLine(PT p, PT p1, PT p2) {
  return fabs(vect(p, p1, p2)) / dis(p1, p2);
}
double disToSeg(PT p, PT p1, PT p2) {
  if (scal(p1, p, p2) < 0) return dis(p, p1);
  if (scal(p2, p, p1) < 0) return dis(p, p2);
  return disToLine(p, p1, p2);
}
bool onLine(PT p, PT p1, PT p2) { return fabs(vect(p1 - p, p2 - p)) < EPS; }
bool onSeg(PT p, PT p1, PT p2) {
  if (!onLine(p, p1, p2)) return 0;
  return (p1.x - p.x) * (p2.x - p.x) < EPS && (p1.y - p.y) * (p2.y - p.y) < EPS;
}
PT projection(PT a, PT b, PT p) {
  return a + (a - b) * (scal(p - a, a - b) / dis2(a - b));
}
int posLineLine(PT p1, PT p2, PT p3, PT p4, PT& p) {
  double s1 = vect(p1, p2, p3), s2 = vect(p1, p2, p4);
  if (fabs(s1 - s2) < EPS) {
    if (fabs(s1) < EPS) return 2;
    return 0;
  }
  p = p3 + (p4 - p3) * s1 / (s1 - s2);
  return 1;
}
const int N = 110;
PT p[N], q[N], r[N];
;
int main() {
  int n;
  cin >> n;
  if (n < 5) return puts("No solution"), 0;
  double a = PI / n;
  double d = 20 / tan(a);
  double step;
  if (n > 20)
    step = d / n / 300;
  else
    step = d / n / 50;
  for (int i = 0; i < n; i++) {
    double ang = a * 2 * i + 0.001;
    p[i] = PT(cos(ang), sin(ang)) * (d + step * i);
    q[i] = p[i] + PT(-sin(ang), cos(ang));
  }
  for (int i = 0; i < n; i++) {
    posLineLine(p[i], q[i], p[(i + 1) % n], q[(i + 1) % n], r[i]);
  }
  double ma = 0, mi = INF;
  for (int i = 0; i < n; i++) {
    double d = dis(r[i], r[(i + 1) % n]);
    ma = max(ma, d);
    mi = min(mi, d);
  }
  assert(ma / mi < 999);
  double mul = 999 / ma;
  for (int i = 0; i < n; i++) r[i] = r[i] * mul;
  double di[N];
  for (int i = 0; i < n; i++) {
    di[i] = dis(r[i], r[(i + 1) % n]);
    double ang =
        getAngle(r[(i + 1) % n] - r[i]) - getAngle(r[i] - r[(i - 1 + n) % n]);
    while (ang < 0) ang += 2 * PI;
    while (ang > 2 * PI) ang -= 2 * PI;
    assert(fabs(ang - PI * 2 / n) < EPS);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < i; j++) {
      assert(fabs(di[i] - di[j]) > 0.001);
    }
  }
  for (int i = 0; i < n; i++) {
    printf("%.10lf %.10lf\n", r[i].x, r[i].y);
  }
  return 0;
}