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 inf = ~0U >> 1;
const long long INF = ~0ULL >> 1;
const double pi = acos(-1.0);
template <class T>
inline void read(T &n) {
  char c;
  for (c = getchar(); !(c >= '0' && c <= '9'); c = getchar())
    ;
  n = c - '0';
  for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0';
}
int Pow(int base, int n, int mo) {
  if (n == 0) return 1;
  if (n == 1) return base % mo;
  int tmp = Pow(base, n >> 1, mo);
  tmp = (long long)tmp * tmp % mo;
  if (n & 1) tmp = (long long)tmp * base % mo;
  return tmp;
}
struct point {
  double x, y;
  point() {}
  point(double _x, double _y) : x(_x), y(_y) {}
};
struct line {
  double k, b;
  int ver;
  line() {}
  line(double _k, double _b, int _ver = 0) : k(_k), b(_b), ver(_ver) {}
};
double R = 1000;
inline line gen_line(const point A, const point B) {
  if (fabs(A.x - B.x) < 1e-9) return line(0, A.x, 1);
  double k = (B.y - A.y) / (B.x - A.x);
  double b = A.y - A.x * k;
  return line(k, b);
}
inline point intersect(line A, line B) {
  if (B.ver) swap(A, B);
  if (A.ver) return point(A.b, B.k * A.b + B.b);
  double x = (B.b - A.b) / (A.k - B.k);
  double y = A.k * x + A.b;
  return point(x, y);
}
template <class T>
T sqr(const T &x) {
  return x * x;
}
inline double dis(const point &A, const point &B) {
  return sqrt(sqr(A.x - B.x) + sqr(A.y - B.y));
}
int n;
point p[200];
inline int pre(int x) { return (x == 1) ? n : (x - 1); }
inline int suc(int x) { return (x == n) ? 1 : (x + 1); }
vector<double> Len;
map<double, int> edge;
int timestamp = 0;
int check() {
  ++timestamp;
  edge.clear();
  Len.clear();
  double std_angle = -1;
  for (int i = (1); i <= (n); ++i) {
    double length = dis(p[i], p[i + 1]);
    int len = int(length * 100 + 0.5);
    if (edge[len] || edge[len - 1] || edge[len + 1]) return 0;
    edge[len] = 1;
    point A = p[suc(i)], B = p[pre(i)], C = p[i];
    double angle =
        acos(((A.x - C.x) * (B.x - C.x) + (A.y - C.y) * (B.y - C.y)) /
             (dis(A, C) * dis(B, C)));
    Len.push_back(length);
    if (std_angle < 1)
      std_angle = angle;
    else if (fabs(std_angle - angle) > 1e-3)
      return 0;
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  R = n * 45;
  if (n <= 4) {
    puts("No solution");
    return 0;
  }
  for (int i = (1); i <= (n); ++i) {
    double angle = (double)(i - 1) / n * (2 * pi);
    p[i].x = R * cos(angle);
    p[i].y = R * sin(angle);
  }
  while (!check()) {
    for (int i = (1); i <= (n); ++i) {
      line A = gen_line(p[pre(i)], p[i]);
      line B = gen_line(p[suc(i)], p[suc(suc(i))]);
      line C = gen_line(p[i], p[suc(i)]);
      C.b += (rand() % 300) / 100.;
      p[i] = intersect(A, C);
      p[suc(i)] = intersect(B, C);
    }
  }
  for (int i = (1); i <= (n); ++i) printf("%.6lf %.6lf\n", p[i].x, p[i].y);
  return 0;
}