#include<cstdio>

#include<cstring>

#include<cmath>

#include<queue>

#include<algorithm>

using namespace std;

typedef long long LL;

const int maxn = ;

int n, T;

int s[maxn], flag[maxn], tmp[maxn];

int u[], v[];

LL a[maxn], b[maxn];

void ls()

{

    int cnt = ;

    for (int i = ; i <= n; i++)

    {

        if (flag[i] == )

            tmp[cnt++] = i;

    }

}

int Find(int a)

{

    for (int i = a; i <= n; i++)

        if (flag[i] == ) return i;

    for (int i = ; i <= a; i++)

        if (flag[i] == ) return i;

}

int f2(int a)

{

    for (int i = ;; i++)

        if (tmp[i] == a) return i;

}

void egcd(LL a, LL b, LL&d, LL&x, LL&y)

{

    if (!b) { d = a, x = , y = ; }

    else

    {

        egcd(b, a%b, d, y, x);

        y -= x*(a / b);

    }

}

LL lmes() {

    LL M = a[], R = b[], x, y, d;

    for (int i = ; i <= n; i++) {

        egcd(M, a[i], d, x, y);

        if ((b[i] - R) % d) return -;

        x = (b[i] - R) / d*x % (a[i] / d);

        R += x*M;

        M = M / d*a[i];

        R %= M;

    }

    return (R + M) % M ? (R + M) % M : M;

}

void exgcd(LL a, LL b, LL &d, LL &x, LL &y)

{

    if (b == )

        d = a, x = , y = ;

    else

    {

        exgcd(b, a%b, d, y, x);

        y -= x * (a / b);

    }

}

LL china(LL n, LL m[], LL a[])

{

    LL aa = a[];

    LL mm = m[];

    for (int i = ; i<n; i++)

    {

        LL sub = (a[i] - aa);

        LL d, x, y;

        exgcd(mm, m[i], d, x, y);

        if (sub % d) return -;

        LL new_m = m[i] / d;

        new_m = (sub / d*x%new_m + new_m) % new_m;

        aa = mm*new_m + aa;

        mm = mm*m[i] / d;

    }

    aa = (aa + mm) % mm;

    return aa ? aa : mm;

}

int main()

{

    scanf("%d", &T);

    while (T--)

    {

        scanf("%d", &n);

        for (int i = ; i <= n; i++)

        {

            int ai; scanf("%d", &ai);

            s[ai] = i;

        }

        memset(flag, , sizeof flag);

        u[] = , v[] = s[];

        if (v[] == n) v[] = ;

        flag[s[]] = ;

        for (int i = ; i <= n; i++)

        {

            int st, en;

            ls();

            st = Find(s[i - ]), st = f2(st), st = st - ;

            if (st == -) st = n - i;

            en = f2(s[i]), u[i] = st;

            v[i] = en; if (v[i] == n - i+) v[i] = ;

            flag[s[i]] = ;

        }

        /*

        for (int i = 1; i <= n; i++)

            printf("%d %d\n", u[i], v[i]);

        */

        //接下来就是求解最小的k，满足所有式子

        //k%(n-i+1)=v[i]-u[i]%(n-i+1)

        for (int i = ; i <= n; i++)

        {

            a[i] = (LL)(n - i + );

            b[i] = (LL)(v[i] - u[i] % (n - i + ));

        }

        LL k = china(n, a + , b + );

        if (k == -) printf("Creation August is a SB!\n");

        else printf("%lld\n", k);

    }

    return ;

}