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;
using namespace std;
struct P {
  double x, y;
  P() {}
  P(double _x, double _y) {
    x = _x;
    y = _y;
  }
  P operator-(const P p1) { return P(x - p1.x, y - p1.y); }
  P operator+(const P p1) { return P(x + p1.x, y + p1.y); }
  P operator/(double r) { return P(x / r, y / r); }
  P operator*(double r) { return P(x * r, y * r); }
  bool operator==(const P p1) const {
    return fabs(x - p1.x) < 1e-8 && fabs(y - p1.y) < 1e-8;
  }
  bool operator!=(const P p1) const {
    return fabs(x - p1.x) > 1e-8 || fabs(y - p1.y) > 1e-8;
  }
  bool operator<(const P p1) const {
    return y < p1.y - 1e-8 || y < p1.y + 1e-8 && x < p1.x;
  }
};
double cpr(const P &a, const P &b, const P &c) {
  return (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
}
double dpr(const P &a, const P &b, const P &c) {
  return (b.x - a.x) * (c.x - a.x) + (b.y - a.y) * (c.y - a.y);
}
double cpr(const P &a, const P &b) { return a.x * b.y - a.y * b.x; }
double dpr(const P &a, const P &b) { return a.x * b.x + a.y * b.y; }
double vlen(const P &a) { return sqrt(a.x * a.x + a.y * a.y); }
double dis(const P &a, const P &b) {
  return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}
bool over(const P &a, const P &b, const P &c, const P &d) {
  double p, q;
  p = cpr(a, b, c);
  q = cpr(a, b, d);
  return p > 1e-8 && q < -1e-8 || p < -1e-8 && q > 1e-8;
}
bool cross(const P &a, const P &b, const P &c, const P &d) {
  double p, q;
  p = cpr(a, b, c);
  q = cpr(a, b, d);
  if (!(p > 1e-8 && q < -1e-8 || p < -1e-8 && q > 1e-8)) return 0;
  p = cpr(c, d, a);
  q = cpr(c, d, b);
  return p > 1e-8 && q < -1e-8 || p < -1e-8 && q > 1e-8;
}
P its(const P &a, const P &b, const P &c, const P &d) {
  P ret = a;
  double t = ((c.x - a.x) * (d.y - c.y) - (c.y - a.y) * (d.x - c.x)) /
             ((b.x - a.x) * (d.y - c.y) - (b.y - a.y) * (d.x - c.x));
  ret.x += (b.x - a.x) * t;
  ret.y += (b.y - a.y) * t;
  return ret;
}
double angsum(P p[], int n) {
  double ret = 0, tmp;
  for (int i = 0; i < n; i++) {
    P &A = p[i], &B = p[(i + 1) % n], &C = p[(i + 2) % n];
    tmp = 3.14159265358979323846 - acos(dpr(B, A, C) / dis(A, B) / dis(B, C));
    if (cpr(A, B, C) < 0) tmp = -tmp;
    ret += tmp;
  }
  return ret;
}
int is_convex(P p[], int n) {
  int s[3] = {1, 1, 1};
  for (int i = 0; i < n && s[1] | s[2]; i++)
    s[((cpr(p[i], p[(i + 1) % n], p[(i + 2) % n])) > 1e-8
           ? 1
           : ((cpr(p[i], p[(i + 1) % n], p[(i + 2) % n])) < -1e-8 ? 2 : 0))] =
        0;
  return s[1] | s[2];
}
int is_convex_v2(P p[], int n) {
  int s[3] = {1, 1, 1};
  for (int i = 0; i < n && s[0] && s[1] | s[2]; i++)
    s[((cpr(p[i], p[(i + 1) % n], p[(i + 2) % n])) > 1e-8
           ? 1
           : ((cpr(p[i], p[(i + 1) % n], p[(i + 2) % n])) < -1e-8 ? 2 : 0))] =
        0;
  return s[0] && s[1] | s[2];
}
int inside_convex(P q, P p[], int n) {
  int s[3] = {1, 1, 1};
  for (int i = 0; i < n && s[1] | s[2]; i++)
    s[((cpr(p[i], p[(i + 1) % n], q)) > 1e-8
           ? 1
           : ((cpr(p[i], p[(i + 1) % n], q)) < -1e-8 ? 2 : 0))] = 0;
  return s[1] | s[2];
}
int inside_convex_v2(P q, P p[], int n) {
  int s[3] = {1, 1, 1};
  for (int i = 0; i < n && s[0] && s[1] | s[2]; i++)
    s[((cpr(p[i], p[(i + 1) % n], q)) > 1e-8
           ? 1
           : ((cpr(p[i], p[(i + 1) % n], q)) < -1e-8 ? 2 : 0))] = 0;
  return s[0] && s[1] | s[2];
}
P barycenter(int n, P *p) {
  P ret(0, 0), t;
  double t1 = 0, t2;
  for (int i = 1; i < n - 1; i++) {
    if (fabs(t2 = cpr(p[i + 1], p[0], p[i])) > 1e-8) {
      t.x = (p[0].x + p[i].x + p[i + 1].x) / 3.0;
      t.y = (p[0].y + p[i].y + p[i + 1].y) / 3.0;
      ret.x += t.x * t2;
      ret.y += t.y * t2;
      t1 += t2;
    }
  }
  if (fabs(t1) > 1e-8) ret.x /= t1, ret.y /= t1;
  return ret;
}
void half_its(P &a, P &b, P pol[], int &polcnt, P exch[]) {
  int i, p2 = 0;
  bool now, last = cpr(a, b, pol[polcnt - 1]) > -1e-8;
  for (i = 0; i < polcnt; i++) {
    now = cpr(a, b, pol[i]) > -1e-8;
    if (now ^ last)
      exch[p2++] = its(a, b, pol[i], pol[(i + polcnt - 1) % polcnt]);
    if (now) exch[p2++] = pol[i];
    last = now;
  }
  polcnt = p2;
  for (i = 0; i < p2; i++) pol[i] = exch[i];
}
P ptoline(P p, P a, P b) {
  P t = p;
  t.x += a.y - b.y;
  t.y += b.x - a.x;
  return its(p, t, a, b);
}
double disptoline(const P &p, const P &a, const P &b) {
  return fabs(cpr(p, a, b)) / dis(a, b);
}
P ptoseg(P p, P a, P b) {
  P t = p;
  t.x += a.y - b.y;
  t.y += b.x - a.x;
  if (cpr(p, a, t) * cpr(p, b, t) > 1e-8) return dis(p, a) < dis(p, b) ? a : b;
  return its(p, t, a, b);
}
double disptoseg(P p, P a, P b) {
  P t = p;
  t.x += a.y - b.y;
  t.y += b.x - a.x;
  if (cpr(p, a, t) * cpr(p, b, t) > 1e-8) return min(dis(p, a), dis(p, b));
  return fabs(cpr(b, p, a)) / dis(a, b);
}
P rotate(P v, P p, double angle, double scale) {
  P ret = p;
  v.x -= p.x;
  v.y -= p.y;
  p.x = scale * cos(angle);
  p.y = scale * sin(angle);
  ret.x += v.x * p.x - v.y * p.y;
  ret.y += v.x * p.y + v.y * p.x;
  return ret;
}
P circumcenter(P a, P b, P c) {
  P u1, u2, v1, v2;
  u1.x = (a.x + b.x) / 2;
  u1.y = (a.y + b.y) / 2;
  u2.x = u1.x - a.y + b.y;
  u2.y = u1.y + a.x - b.x;
  v1.x = (a.x + c.x) / 2;
  v1.y = (a.y + c.y) / 2;
  v2.x = v1.x - a.y + c.y;
  v2.y = v1.y + a.x - c.x;
  return its(u1, u2, v1, v2);
}
P incenter(P a, P b, P c) {
  P u1, u2, v1, v2;
  double m, n;
  u1 = a;
  m = atan2(b.y - a.y, b.x - a.x);
  n = atan2(c.y - a.y, c.x - a.x);
  u2.x = u1.x + cos((m + n) / 2);
  u2.y = u1.y + sin((m + n) / 2);
  v1 = b;
  m = atan2(a.y - b.y, a.x - b.x);
  n = atan2(c.y - b.y, c.x - b.x);
  v2.x = v1.x + cos((m + n) / 2);
  v2.y = v1.y + sin((m + n) / 2);
  return its(u1, u2, v1, v2);
}
P perpencenter(P a, P b, P c) {
  P u1, u2, v1, v2;
  u1 = c;
  u2.x = u1.x - a.y + b.y;
  u2.y = u1.y + a.x - b.x;
  v1 = b;
  v2.x = v1.x - a.y + c.y;
  v2.y = v1.y + a.x - c.x;
  return its(u1, u2, v1, v2);
}
P barycenter(P a, P b, P c) {
  P ret;
  ret.x = (a.x + b.x + c.x) / 3;
  ret.y = (a.y + b.y + c.y) / 3;
  return ret;
}
P fermentpoint(P a, P b, P c) {
  P u, v;
  double step =
      fabs(a.x) + fabs(a.y) + fabs(b.x) + fabs(b.y) + fabs(c.x) + fabs(c.y);
  int i, j, k;
  u.x = (a.x + b.x + c.x) / 3;
  u.y = (a.y + b.y + c.y) / 3;
  while (step > 1e-10) {
    for (k = 0; k < 10; step /= 2, k++)
      for (i = -1; i <= 1; i++)
        for (j = -1; j <= 1; j++) {
          v.x = u.x + step * i;
          v.y = u.y + step * j;
          if (dis(u, a) + dis(u, b) + dis(u, c) >
              dis(v, a) + dis(v, b) + dis(v, c))
            u = v;
        }
  }
  return u;
}
void intersection_line_circle(P c, double r, P l1, P l2, P &p1, P &p2) {
  P p = c;
  p.x += l1.y - l2.y;
  p.y += l2.x - l1.x;
  p = its(p, c, l1, l2);
  double d = dis(p, c), t = sqrt(r * r - d * d) / dis(l1, l2);
  p2.x = p.x + (l2.x - l1.x) * t;
  p2.y = p.y + (l2.y - l1.y) * t;
  p1.x = p.x - (l2.x - l1.x) * t;
  p1.y = p.y - (l2.y - l1.y) * t;
}
void intersection_circle_circle(P c1, double r1, P c2, double r2, P &p1,
                                P &p2) {
  P u, v;
  double t = (1 + (r1 * r1 - r2 * r2) / dis(c1, c2) / dis(c1, c2)) / 2;
  u.x = c1.x + (c2.x - c1.x) * t;
  u.y = c1.y + (c2.y - c1.y) * t;
  v.x = u.x + c1.y - c2.y;
  v.y = u.y - c1.x + c2.x;
  intersection_line_circle(c1, r1, u, v, p1, p2);
}
void find_tp(double a, double b, double c, double &ang1, double &ang2) {
  double v1, v2;
  v1 = fabs(c) > 1e-8 ? atan2(b * c, a * c) : atan2(b, a);
  v2 = acos(fabs(c) / sqrt(a * a + b * b));
  ang1 = v1 - v2;
  ang2 = v1 + v2;
}
void tangent1(P c1, double r1, P c2, double r2, double &t1, double &t2) {
  find_tp(c2.x - c1.x, c2.y - c1.y, r1 - r2, t1, t2);
}
void tangent2(P c1, double r1, P c2, double r2, double &t1, double &t2) {
  find_tp(c2.x - c1.x, c2.y - c1.y, r1 + r2, t1, t2);
}
void make_ch(P *p, P *f, int n, int &top) {
  top = 0;
  sort(p, p + n);
  for (int i = 0; i < 2 * n - 1; i++) {
    int j = (i < n) ? i : 2 * (n - 1) - i;
    while (top > 1 && cpr(f[top - 2], f[top - 1], p[j]) < -1e-8) top--;
    f[top++] = p[j];
  }
  top--;
}
void make_ch(P *p, int *s, int n, int &top) {
  top = 0;
  sort(p, p + n);
  for (int i = 0; i < 2 * n - 1; i++) {
    int j = (i < n) ? i : 2 * (n - 1) - i;
    while (top > 1 && cpr(p[s[top - 2]], p[s[top - 1]], p[j]) < -1e-8) top--;
    s[top++] = j;
  }
  top--;
}
int main() {
  int n;
  double mi = 1e60;
  double mil = 1e60;
  double mal = 1e0;
  while (cin >> n) {
    vector<P> v;
    double apf = 2 * 3.14159265358979323846 / n;
    double x = 0.0;
    const double TT = 800;
    for (int i = 0; i < n; i++) {
      P p;
      p.x = TT * cos(x);
      p.y = TT * sin(x);
      x += apf;
      v.push_back(p);
    }
    const double T = 50;
    vector<P> vp;
    for (int i = 0; i < n; i++) {
      P mid = (v[i] + v[(i + 1) % n]) / 2.0;
      mid = mid / vlen(mid);
      P p1 = v[i] + mid * i / T;
      P p2 = v[(i + 1) % n] + mid * i / T;
      mid = (v[(i + 1) % n] + v[(i + 2) % n]) / 2.0;
      mid = mid / vlen(mid);
      P p3 = v[(i + 1) % n] + mid * ((i + 1) % n) / T;
      P p4 = v[(i + 2) % n] + mid * ((i + 1) % n) / T;
      vp.push_back(its(p1, p2, p3, p4));
    }
    bool suc = true;
    for (int i = 0; i < ((int)(vp.size())); i++) {
      double len1 = dis(vp[i], vp[(i + 1) % n]);
      mil = min(mil, len1);
      mal = max(mal, len1);
      for (int j = 0; j < i; j++) {
        double len2 = dis(vp[j], vp[j + 1]);
        if ((fabs(len1 - len2) < 1e-8)) suc = false;
        mi = min(mi, fabs(len1 - len2));
      }
    }
    if (suc) {
      for (int i = 0; i < n; i++) printf("%.10lf %.10lf\n", vp[i].x, vp[i].y);
    } else {
      puts("No solution");
    }
  }
}