CSL Points Range

In the Chinese Super League (CSL), the rules for each match are as follows:

• A win (when your score is strictly greater than the opponent's) gives you $3$ points,
• A draw (when both scores are equal) gives you $1$ point,
• A loss (when your score is strictly less than the opponent's) gives you $0$ points.

You are given that you played $N$ matches in total, scoring $S$ goals and conceding $T$ goals overall. In each match, you can decide the result, but the overall totals must be exactly $S$ and $T$. Note that for a win, you must score at least one more goal than your opponent, and for a loss, you must score at least one goal less than your opponent. In a draw, the scores are equal. In every match you are free to add extra goals to both teams as long as the relative outcome does not change.

Using a base configuration for each match:

• Win: assign $(1, 0)$ (difference $+1$).
• Draw: assign $(0, 0)$ (difference $0$).
• Loss: assign $(0, 1)$ (difference $-1$).

Let the number of wins, draws and losses be $x$, $y$, and $z$ respectively. Then we have:

$$\begin{array}{l} x+y+z = N,\\ \text{Total points} = 3x + y,\\ \text{And the base difference: }x - z = S - T = D. \end{array}$$

Extra goals can be added equally to both teams in a match so as not to change the outcome. This means that given $x$, $y$, and $z$ satisfying the above with

$$x - z = D,$$

a valid assignment exists if and only if the following conditions can be met:

1. $x \le S$ (since in wins, the minimum goals scored is $1$ each),
2. $z \le T$, (since in losses, the minimum conceded goals is $1$ each), and
3. $x+z \le N$ (since draws make up the remainder).

We can rewrite $z = x - D$. Then the feasibility domain for $x$ is:

$$\max(0, D) \le x \le \min\Big(S, T+D, \;\frac{N+D}{2}\Big). $$

Since the total points equal

$$P = 3x + y = 3x + (N - x - z) = 3x + N - x - (x-D) = x + N + D, $$

the total points is a linear function of $x$. Hence, the minimum total points correspond to the minimum possible valid $x$ and the maximum total points correspond to the maximum possible valid $x$.

Let:

$$x_{min} = \max(0, D) \quad \text{and} \quad x_{max} = \min\Big(S, T+D, \;\left\lfloor\frac{N+D}{2}\right\rfloor\Big). $$

Then the minimum and maximum points that can be achieved are:

• Minimum Points: $P_{min} = x_{min} + N + D$,
• Maximum Points: $P_{max} = x_{max} + N + D$.

Your task is to compute these two values.

inputFormat

The input consists of a single line containing three integers separated by spaces:

• $N$ ($1 \leq N \leq 10^9$): the number of matches,
• $S$ ($0 \leq S \leq 10^9$): the total goals scored,
• $T$ ($0 \leq T \leq 10^9$): the total goals conceded.

It is guaranteed that a valid distribution exists (note that $S-T$ cannot exceed $N$ or be less than $-N$).

outputFormat

Output two integers separated by a space: the maximum possible total points and the minimum possible total points respectively.

sample

3 4 2

7 7