Codeforces1313 B. Different Rules（找规律）

Description：

Nikolay has only recently started in competitive programming, but already qualified to the finals of one prestigious olympiad. There going to be n participants, one of whom is Nikolay. Like any good olympiad, it consists of two rounds. Tired of the traditional rules, in which the participant who solved the largest number of problems wins, the organizers came up with different rules.

Suppose in the first round participant A took $x-th$ place and in the second round — $y-th$ place. Then the total score of the participant A is sum $x+y$. The overall place of the participant A is the number of participants (including A) having their total score less than or equal to the total score of A. Note, that some participants may end up having a common overall place. It is also important to note, that in both the first and the second round there were no two participants tying at a common place. In other words, for every i from 1 to n exactly one participant took i-th place in first round and exactly one participant took $i-th$ place in second round.

Right after the end of the Olympiad, Nikolay was informed that he got x-th place in first round and y-th place in the second round. Nikolay doesn’t know the results of other participants, yet he wonders what is the minimum and maximum place he can take, if we consider the most favorable and unfavorable outcome for him. Please help Nikolay to find the answer to this question.

Input

The first line contains an integer $t (1≤t≤100)$ — the number of test cases to solve.

Each of the following t lines contains integers $n, x, y (1≤n≤10^9, 1≤x,y≤n)$ — the number of participants in the olympiad, the place that Nikolay took in the first round and the place that Nikolay took in the second round.

Output

Print two integers — the minimum and maximum possible overall place Nikolay could take.

Examples

input

1
5 1 3

output

1 3

input

1
6 3 4

output

2 6

Note

Explanation for the first example:

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Suppose there were $5$ participants A-E. Let’s denote Nikolay as A. The the most favorable results for Nikolay could look as follows:

However, the results of the Olympiad could also look like this:

In the first case Nikolay would have taken first place, and in the second — third place.

题意：

$n$ 个人参加比赛一共两场，给出一个人的两场比赛排名为 $x,y$，问这个 人的最高排名和最低排名。
全部人的两场排名之和范围是 $2~2n$ 。

• 先求最高排名，如果这个人 $x+y<n-1$ ,那么他的总排名肯定可以最高为1，如果大于的话那最低排名就只能是 $n$。
• 最低排名的话，如果 $x+y<n+1$，那就是 $x+y-1$,否则就是 $n$。
  这个很好推啊，不知道咋卡了怎么多人，我一开始写了好几个判断，后来化简一下就是两个式子。

AC代码：

const int N = 25;
int n, m;
int t;
int a, b, c;
int x, y;
int pos;
int res, tmp, ans1, ans2;
int main()
{
    sd(t);
    while (t--)
    {
        sddd(n, x, y);
        res = x + y;
        ans1 = min(max(res + 1 - n, 1), n);
        ans2 = min(res - 1, n);
        pdd(ans1, ans2);
    }
    return 0;
}