A. Riptide - 技术栈

time limit per test

1 second

memory limit per test

256 megabytes

Alice, Bob, and Charlie are playing a game with tokens. They start with a, b, and c tokens, respectively.

The game is played in rounds. Before the beginning of each round, they check the number of tokens everyone has:

If any two players have the exact same number of tokens, the game immediately ends.
Otherwise, the round begins, all three players have a strictly different number of tokens. The player with the strictly most tokens gives exactly 1 token to the player with the strictly fewest tokens.

Given the starting tokens a, b, and c, determine exactly how many rounds the game will last before it ends.

Input

The first line contains a single integer t (1≤t≤103) --- the number of test cases.

Each test case consists of a single line containing three integers a, b, and c (1≤a,b,c≤10).

Output

For each test case, output a single integer --- the number of rounds the game will last before it ends.

Example

Input

Copy

复制代码

1 2 3

4 6 1

3 3 7

1 7 10

6 1 9

1 1 1

Output

Copy

复制代码

Note

In the first test case:

No two players have the same number of tokens.
Charlie has the most tokens (3 tokens), and Alice has the fewest tokens (1 token). Therefore, Charlie gives Alice a token.
Now, Alice has 2 tokens, Bob has 2 tokens, and Charlie has 2 tokens. Since there are two players (or more) with the same number of tokens, the game ends.

The game ended after 1 round, so the answer is 1.

In the second test case, the game is played as follows:

Bob gives Charlie a token, now Alice has 4 tokens, Bob has 5 tokens, and Charlie has 2 tokens.
Bob gives Charlie a token, now Alice has 4 tokens, Bob has 4 tokens, and Charlie has 3 tokens. Since two players have the same number of tokens, the game ends.

The game lasted 2 rounds.

In the third test case, two players already have the same number of tokens. So the answer is 0 since no rounds were played.

解题说明：水题，找到差值最小的两个数即可。

cpp 复制代码

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) 
    {
        int a, b, c, count = 0;
        scanf("%d%d%d", &a, &b, &c);
        count = abs(a - b);
        if (count > abs(b - c)) 
        {
            count = abs(b - c);
        }
        if (count > abs(c - a))
        {
            count = abs(c - a);
        }
        printf("%d\n", count);
    }
    return 0;
}