Polycarp is playing a new computer game. This game has n stones in a row. The stone on the position i has integer power a_i. 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 = 5 and 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]; 
  * Destroy the rightmost stone. After this move a = [5, 4, 3]; 
  * Destroy the leftmost stone. After this move 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]; 
  * Destroy the leftmost stone. After this move 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 t (1 ≤ t ≤ 100). Then t test cases follow.

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

The second line contains n distinct integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 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


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


2
4
5
3
2
// Problem: A. Stone Game
// Contest: Codeforces - Codeforces Round #725 (Div. 3)
// URL: https://codeforces.com/contest/1538/problem/A
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>   //blitztage
using namespace std;
 
//for loop macros
#define fo(i,n) for(int i = 0; i < n; i++)
#define foL(i,n) for(long long i = 0; i < n; i++)
#define foa(i,k,n) for(int i = k; i < n; i++)
#define fob(i,k,n) for(int i = k; i >= n; i--)
#define foaL(i,k,n) for(long long i = k; i < n; i++)
#define fobL(i,k,n) for(long long i = k; i >= n; i--)
 
constexpr int bits(int x) {return x == 0 ? 0 : 31-__builtin_clz(x);}
 
//template -> AnandOza && bqi343 Github
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define F first
#define S second
#define resz resize
 
#define sz(x) int((x).size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define trav(a, x) for (auto &a : x)
 
#define L1(u, ...) [&](auto &&u) { return __VA_ARGS__; }
#define L2(u, v, ...) [&](auto &&u, auto &&v) { return __VA_ARGS__; }
 
#define sort_by(x, y) sort(all(x), [&](const auto &l, const auto &r) { return y; })
 
using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vd = vector<double>;
using vs = vector<string>;
 
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pdd = pair<double, double>;
 
using vpii = vector<pii>;
using vpll = vector<pll>;
using vpdd = vector<pdd>;
 
template <typename T> void ckmin(T &a, const T &b) { a = min(a, b); }
template <typename T> void ckmax(T &a, const T &b) { a = max(a, b); }
 
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
 
namespace __input {
    template <class T1, class T2> void re(pair<T1, T2> &p);
    template <class T> void re(vector<T> &a);
    template <class T, size_t SZ> void re(array<T, SZ> &a);
 
    template <class T> void re(T &x) { cin >> x; }
    void re(double &x) { string t; re(t); x = stod(t); }
    template <class Arg, class... Args> void re(Arg &first, Args &...rest) { re(first); re(rest...); }
 
    template <class T1, class T2> void re(pair<T1, T2> &p) { re(p.f, p.s); }
    template <class T> void re(vector<T> &a) { for (int i = 0; i < sz(a); i++) re(a[i]); }
    template <class T, size_t SZ> void re(array<T, SZ> &a) { for (int i = 0; i < SZ; i++) re(a[i]); }
}
using namespace __input;
 
namespace __output {
    template <typename T> struct is_outputtable { template <typename C> static constexpr decltype(declval<ostream &>() << declval<const C &>(), bool()) test(int) { return true; } template <typename C> static constexpr bool test(...) { return false; } static constexpr bool value = test<T>(int()); };
    template <class T, typename V = decltype(declval<const T &>().begin()), typename S = typename enable_if<!is_outputtable<T>::value, bool>::type> void pr(const T &x);
 
    template <class T, typename V = decltype(declval<ostream &>() << declval<const T &>())> void pr(const T &x) { cout << x; }
    template <class T1, class T2> void pr(const pair<T1, T2> &x);
    template <class Arg, class... Args> void pr(const Arg &first, const Args &...rest) { pr(first); pr(rest...); }
 
    template <class T, bool pretty = true> void prContain(const T &x) { if (pretty) pr("{"); bool fst = 1; for (const auto &a : x) pr(!fst ? pretty ? ", " : " " : "", a), fst = 0; if (pretty) pr("}"); }
 
    template <class T> void pc(const T &x) { prContain<T, false>(x); pr("\n"); }
    template <class T1, class T2> void pr(const pair<T1, T2> &x) { pr("{", x.f, ", ", x.s, "}"); }
    template <class T, typename V, typename S> void pr(const T &x) { prContain(x); }
 
    void ps() { pr("\n"); }
    template <class Arg> void ps(const Arg &first) { pr(first); ps(); }
    template <class Arg, class... Args> void ps(const Arg &first, const Args &...rest) { pr(first, " "); ps(rest...); }
}
using namespace __output;
 
#define __pn(x) pr(#x, " = ")
#ifdef ANAND_LOCAL
#define pd(...) pr("\033[1;31m"), __pn((__VA_ARGS__)), ps(__VA_ARGS__), pr("\033[0m"), cout << flush
#else
#define pd(...)
#endif
 
namespace __algorithm {
    template <typename T> void dedup(vector<T> &v) { sort(all(v)); v.erase(unique(all(v)), v.end()); }
    template <typename T> typename vector<T>::const_iterator find(const vector<T> &v, const T &x) { auto it = lower_bound(all(v), x); return it != v.end() && *it == x ? it : v.end(); }
    template <typename T> size_t index(const vector<T> &v, const T &x) { auto it = find(v, x); assert(it != v.end() && *it == x); return it - v.begin(); }
    template <typename I> struct _reversed_struct { I &v_; explicit _reversed_struct(I &v) : v_{v} {} typename I::reverse_iterator begin() const { return v_.rbegin(); } typename I::reverse_iterator end() const { return v_.rend(); } };
    template <typename I> _reversed_struct<I> reversed(I &v) { return _reversed_struct<I>(v); }
}
using namespace __algorithm;
 
namespace __io {
    void setIO() { ios_base::sync_with_stdio(0); cin.tie(0); cout << setprecision(15); }
}
using namespace __io;
 
template<typename T_vector>
void outvec(const T_vector &v, bool add_one = false, int start = -1, int end = -1) {
    if (start < 0) start = 0;
    if (end < 0) end = int(v.size());
 
    for (int i = start; i < end; i++)
        cout << v[i] + (add_one ? 1 : 0) << (i < end - 1 ? ' ' : '\n');
}
 
//#define int long long
 
#define noo {ps("NO");return;}
#define yess {ps("YES");return;}
 
#define Multitests 1
 
 
void solve(){
	int n;cin >> n;
	vi a(n); re(a);
	vi b = a;
	sort(all(b));
	int x = b[0] ; int y = b[n-1];
	int xx = -1; int yy = -1;
	fo(i,n){
		if(a[i] == x) xx = i;
		if(a[i] == y) yy = i;
	}
	// i from left;; n-1-i from right
	ps(min({xx+n-1-yy+2,yy+n-1-xx+2,max(xx+1,yy+1),max(n-1-yy+1,n-1-xx+1)}));
	// ps(xx+n-1-yy,yy+n-1-xx,max(xx,yy),max(n-1-yy,n-1-xx));
}
 
 
int32_t main() {
    setIO();
	#if Multitests
		int t; cin >> t; while(t--) solve();
	#else 
		solve();
	#endif
    return 0;
}