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 Get() {
  int res = 0, q = 1;
  char ch = getchar();
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  if (ch == '-') q = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') res = res * 10 + ch - '0', ch = getchar();
  return res * q;
}
const double eps = 1e-3, pi = M_PI;
const int oo = (int)2e9, mod = (int)1e9 + 7;
const long long INF = (long long)1e17;
struct vec {
  double x, y;
};
struct line {
  vec p, t;
};
vec operator+(const vec &a, const vec &b) {
  return (vec){a.x + b.x, a.y + b.y};
}
vec operator-(const vec &a, const vec &b) {
  return (vec){a.x - b.x, a.y - b.y};
}
vec operator*(const double &a, const vec &b) { return (vec){a * b.x, a * b.y}; }
double operator/(const vec &a, const vec &b) {
  return fabs(b.x) > 0 ? a.x / b.x : a.y / b.y;
}
double operator*(const vec &a, const vec &b) { return a.x * b.y - a.y * b.x; }
double operator^(const vec &a, const vec &b) { return a.x * b.x + a.y * b.y; }
bool operator==(const vec &a, const vec &b) { return a.x == b.x && a.y == b.y; }
inline bool Equal(double a, double b) { return fabs(a - b) < eps; }
inline double Dist(vec a) { return sqrt(((a.x) * (a.x)) + ((a.y) * (a.y))); }
inline vec Line_Cross(line a, line b) {
  vec P = b.p - a.p;
  double t = (P * b.t) / (a.t * b.t);
  return a.p + t * a.t;
}
vec Rot(vec x, double y) {
  vec d;
  d.x = x.x * cos(y) - x.y * sin(y);
  d.y = x.y * cos(y) + x.x * sin(y);
  return d;
}
double Random() { return 0.95 + (double)(rand() / 5 * 2) / (double)RAND_MAX; }
const int N = 100;
line a[N + 10];
int n;
double ang, l[N + 10];
int main() {
  srand(time(NULL));
  scanf(
      "%d"
      "\n",
      &n);
  if (n <= 4) return printf("No solution\n"), 0;
  ang = 2 * pi / n;
  a[1].t = (vec){400, 0}, a[1].p = (vec){0, 0};
  for (int i = (2), end = (n); i <= end; ++i) {
    a[i].p = a[i - 1].p + a[i - 1].t;
    a[i].t = Rot(a[i - 1].t, ang);
  }
  for (;;) {
    bool ext = 1;
    for (int i = (1), end = (n); i <= end; ++i) {
      l[i] = Dist(a[i].t);
      bool flag = 0;
      for (int j = (1), end = (i - 1); j <= end; ++j)
        if (Equal(l[i], l[j])) {
          flag = 1;
          break;
        }
      if (!flag) continue;
      ext = 0;
      int nx = i % n + 1, nt = nx % n + 1;
      double p = Random();
      vec cr;
      for (;; p = Random()) {
        if (fabs(p - 1) < eps) continue;
        vec np = a[i].p + p * a[i].t;
        line next = (line){np, a[nx].t};
        cr = Line_Cross(next, a[nt]);
        vec dt = cr - np;
        if ((dt / a[nx].t) < eps) continue;
        a[i].t = p * a[i].t, a[nx].p = a[i].p + a[i].t, a[nt].p = cr;
        a[nx].t = a[nt].p - a[nx].p;
        l[i] = Dist(a[i].t), l[nx] = Dist(a[nx].t), l[nt] = Dist(a[nt].t);
        if (l[i] - 1 < eps || 1000 - l[i] < eps) continue;
        if (l[nx] - 1 < eps || 1000 - l[nx] < eps) continue;
        if (l[nt] - 1 < eps || 1000 - l[nt] < eps) continue;
        bool ck = 1;
        for (int j = (1), end = (n); j <= end; ++j)
          if (j != i)
            if (Equal(l[i], l[j])) {
              ck = 0;
              break;
            }
        if (ck) break;
      }
    }
    if (ext) break;
  }
  for (int i = (1), end = (n); i <= end; ++i)
    printf(
        "%.10lf"
        " "
        "%.10lf"
        "\n",
        a[i].p.x, a[i].p.y);
  return 0;
}