Bảng mưa hai năm (Two-Year Rainfall Table)

Source: Problem Solving and Program Design in C

Version: Phuoc Hung OJ Extended

Problem

You are given monthly rainfall for two consecutive years. For year $y\in\{1,2\}$ and month $m\in\{1,2,\ldots,12\}$, let $r_{y,m}$ be the rainfall in millimetres.

The yearly total is

$$T_y=\sum_{m=1}^{12}r_{y,m},$$

and the monthly average is

$$A_y=\frac{T_y}{12}.$$

For every month $m$, define

$$c_m= \begin{cases} -1,&r_{1,m}<r_{2,m},\\ 0,&r_{1,m}=r_{2,m},\\ 1,&r_{1,m}>r_{2,m}. \end{cases}$$

Compute $T_1,A_1,T_2,A_2$ and $c_1,c_2,\ldots,c_{12}$.

Input

The first line contains 12 integers $r_{1,1},\ldots,r_{1,12}$. The second line contains 12 integers $r_{2,1},\ldots,r_{2,12}$.

Output

Line 1 prints $T_1$ and $A_1$. Line 2 prints $T_2$ and $A_2$. Each average must be printed with exactly 6 digits after the decimal point. Line 3 prints $c_1,c_2,\ldots,c_{12}$.

Subtask

Subtask 1 (100 points): $0\le r_{y,m}\le10^6$.

Example

Input

10 20 30 40 50 60 70 80 90 100 110 120
15 15 35 35 55 55 75 75 95 95 115 115

Output

780 65.000000
780 65.000000
-1 1 -1 1 -1 1 -1 1 -1 1 -1 1

Explanation

Both years total $780$ mm and average $65$ mm/month. In month 1, year 1 has less rain so $c_1=-1$; in month 2, year 1 has more rain so $c_2=1$; the same alternating pattern continues.