A. Stone Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp is playing a new computer game. This game has nn stones in a row. The stone on the position ii has integer power aiai. The powers of all stones are distinct.

Each turn Polycarp can destroy either stone on the first position or stone on the last position (in other words, either the leftmost or the rightmost stone). When Polycarp destroys the stone it does not exist any more.

Now, Polycarp wants two achievements. He gets them if he destroys the stone with the least power and the stone with the greatest power. Help Polycarp find out what is the minimum number of moves he should make in order to achieve his goal.

For example, if n=5n=5 and a=[1,5,4,3,2]a=[1,5,4,3,2], then Polycarp could make the following moves:

• Destroy the leftmost stone. After this move a=[5,4,3,2]a=[5,4,3,2];
• Destroy the rightmost stone. After this move a=[5,4,3]a=[5,4,3];
• Destroy the leftmost stone. After this move a=[4,3]a=[4,3]. Polycarp destroyed the stones with the greatest and least power, so he can end the game.

Please note that in the example above, you can complete the game in two steps. For example:

• Destroy the leftmost stone. After this move a=[5,4,3,2]a=[5,4,3,2];
• Destroy the leftmost stone. After this move a=[4,3,2]a=[4,3,2]. Polycarp destroyed the stones with the greatest and least power, so he can end the game.

Input

The first line contains an integer tt (1≤t≤1001≤t≤100). Then tt test cases follow.

The first line of each test case contains one integer nn (2≤n≤1002≤n≤100) — the number of stones.

The second line contains nn distinct integers a1,a2,…,ana1,a2,…,an (1≤ai≤n1≤ai≤n) — the power of the stones.

Output

For each test case, output the minimum number of moves required to destroy the stones with the greatest and the lowest power.

Example

input

Copy

5
5
1 5 4 3 2
8
2 1 3 4 5 6 8 7
8
4 2 3 1 8 6 7 5
4
3 4 2 1
4
2 3 1 4

output

Copy

2
4
5
3
2

解题说明：此题是一道模拟题，先找出最大值最小值所在的位置，然后统计出从最左和最右分别到最大值和最小值的步数。

#include<stdio.h>

int main()
{
	int t, i, j, k, n, a[200], min, max, c;
	scanf("%d", &t);
	for (i = 0; i<t; i++)
	{
		scanf("%d", &n);
		for (j = 0; j<n; j++)
		{
			scanf("%d", &a[j]);
			if (a[j] == 1)
			{
				min = j + 1;
			}
			if (a[j] == n)
			{
				max = j + 1;
			}
		}
		if (min>max)
		{
			k = min;
			min = max;
			max = k;
		}
		c = max;
		if (c > n + 1 - min)
		{
			c = n + 1 - min;
		}
		if (c > min + n + 1 - max)
		{
			c = min + n + 1 - max;
		}
		printf("%d\n", c);
	}
	return 0;
}