Polygon Programming with Ease

There are $N$ Foreign Offices located at distinct points in the plane, placed in order as the vertices of a polygon: $F_1, F_2, \dots, F_N$. On each edge joining two neighbouring Foreign Offices, a single meeting location is placed at the midpoint of that edge. That is, for $i = 1, 2, \dots, N-1$ the meeting location $M_i$ is the midpoint of $F_i$ and $F_{i+1}$, and $M_N$ is the midpoint of $F_N$ and $F_1$.

Given the coordinates of the $N$ meeting locations $M_1, M_2, \dots, M_N$ in that order, determine the coordinates of the $N$ Foreign Offices $F_1, F_2, \dots, F_N$. When $N$ is odd the Foreign Offices are uniquely determined.

The input consists of several test cases and is processed until end of file. Each test case is given on a single line. It starts with an odd integer $N$ ($3 \le N$), followed by $2N$ integers $x_1\ y_1\ x_2\ y_2\ \dots\ x_N\ y_N$, where $(x_i, y_i)$ is the coordinate of the $i$-th meeting location $M_i$, listed in order around the polygon. All coordinates are integers.

For each test case, print one line. First print the integer $N$, then the coordinates of the Foreign Offices $F_1, F_2, \dots, F_N$ in the same order as the meeting locations. Print every coordinate with exactly six digits after the decimal point, and separate all values with single spaces.

The $i$-th meeting location is the midpoint of the $i$-th and $(i+1)$-th Foreign Office (indices are cyclic, so $M_N$ is the midpoint of $F_N$ and $F_1$). Because $N$ is odd, the Foreign Offices can be recovered uniquely: first take the alternating sum $F_1 = M_1 - M_2 + M_3 - \dots + M_N$, then compute the rest with $F_{i+1} = 2M_i - F_i$ for $i = 1, 2, \dots, N-1$.