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 int MM = 1e9 + 7;
const double eps = 1e-8;
const int MAXN = 2e3 + 10;
const int MAXM = 2e6 + 10;
struct Pt {
  double x, y;
};
inline int equal(double p, double q) { return fabs(p - q) < eps; }
bool operator<(const Pt& a, const Pt& b) {
  if (equal(a.x, b.x))
    return a.y < b.y;
  else
    return a.x < b.x;
}
bool operator==(const Pt& a, const Pt& b) {
  return equal(a.x, b.x) && equal(a.y, b.y);
}
inline double operator*(const Pt& a, const Pt& b) {
  return a.x * b.y - a.y * b.x;
}
inline double operator%(const Pt& a, const Pt& b) {
  return a.x * b.x + a.y * b.y;
}
inline Pt operator-(const Pt& a, const Pt& b) {
  return (Pt){a.x - b.x, a.y - b.y};
}
inline Pt operator+(const Pt& a, const Pt& b) {
  return (Pt){a.x + b.x, a.y + b.y};
}
inline Pt operator*(double u, const Pt& a) { return (Pt){a.x * u, a.y * u}; }
double sqr(double x) { return x * x; }
inline double len(const Pt& a) { return sqrt(sqr(a.x) + sqr(a.y)); }
vector<Pt> convex_hull(vector<Pt> u) {
  sort(u.begin(), u.end());
  u.erase(unique(u.begin(), u.end()), u.end());
  if (u.size() < 3) return u;
  vector<Pt> c;
  for (size_t i = 0, o = 1, m = 1; ~i; i += o) {
    while (c.size() > m) {
      Pt a = c.back() - c[c.size() - 2];
      Pt b = c.back() - u[i];
      if (a * b < 0) break;
      c.pop_back();
    }
    c.push_back(u[i]);
    if (i + 1 == u.size()) m = c.size(), o = -1;
  }
  c.pop_back();
  return c;
}
double __cs, __sn;
Pt rotate(const Pt& a) {
  return (Pt){a.x * __cs - a.y * __sn, a.x * __sn + a.y * __cs};
}
Pt rotate(const Pt& a, double s) {
  return (Pt){a.x * cos(s) - a.y * sin(s), a.x * sin(s) + a.y * cos(s)};
}
double f[MAXN];
Pt p[MAXN], nico[MAXN];
int n;
void solve() {
  if (n <= 4) {
    puts("No solution");
    return;
  }
  for (int _ = 1; _ <= 1000000; _++) {
    p[0] = (Pt){252.25252, 0};
    double s = (acos(-1.0) * 2.0 / n);
    for (int i = 1; i < n; i++) {
      p[i] = rotate(p[0], (acos(-1.0) * 2.0 / n * i));
    }
    for (int i = 0; i < n; i++) {
      double x = rand() * 0.5 / RAND_MAX + 0.5;
      f[i] = x;
      p[i] = x * p[i];
    }
    sort(f, f + n);
    double ppp = 1e200;
    for (int i = 1; i < n; i++) ppp = min(ppp, f[i] - f[i - 1]);
    nico[0] = (Pt){0, 0};
    for (int i = 1; i < n - 1; i++) {
      nico[i] = nico[i - 1] + p[i];
    }
    if (nico[n - 2].y <= 0) continue;
    nico[n - 1].y = 0;
    nico[n - 1].x = nico[n - 2].x - (nico[n - 2].y / p[n - 1].y * p[n - 1].x);
    if (nico[n - 1].x >= 0) continue;
    nico[n] = nico[0];
    for (int i = 0; i < n; i++) f[i] = len(nico[i + 1] - nico[i]);
    sort(f, f + n);
    ppp = 1e200;
    for (int i = 1; i < n; i++) ppp = min(ppp, f[i] - f[i - 1]);
    if (f[n - 1] > 999) continue;
    if (ppp >= 5e-3) {
      for (int i = 0; i < n; i++) printf("%.15f %.15f\n", nico[i].x, nico[i].y);
      return;
    }
  }
  puts("No solution");
}
void printans() {}
int main() {
  srand(time(NULL));
  int T = 1;
  while (scanf("%d", &n) != EOF) {
    solve();
    printans();
  }
  return 0;
}