#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <vector>
#include <string>
#define LL long longusing namespace std;int n, s[25];
bool vis[25];
LL rest[25], mmod[25];LL gcd(LL a, LL b)
{LL r;while (b != 0){r = b;b = a%b;a = r;}return a;
}LL lcm(LL a, LL b)
{return a/gcd(a, b)*b;
}//EXGCD
//求解方程ax+by=d,即ax=d mod(b)
//扩展可求逆元
//O(logn)
void exgcd(LL a, LL b, LL &x, LL &y, LL &d)
{if (b == 0){x = 1;y = 0;d = a;}else{exgcd(b, a%b, y, x, d);y -= a/b*x;}
}//中国剩余定理（非互质）
//其中a为除数数组,n为模数数组
LL inv(LL a, LL n)
{LL x, y, d;exgcd(a, n, x, y, d);if (d != 1)return -1;return (x%n + n) % n;
}bool Merge(LL a1, LL n1, LL a2, LL n2, LL &aa, LL &nn)
{LL d = gcd(n1, n2);LL c = a2 - a1;if (c % d)return false;//(n1/d)*k1 = (c/d)(mod(n2/d))c /= d;n1 /= d;n2 /= d;//k1 = (c/d)*(n1/d)^(-1)(mod(n2/d))c *= inv(n1, n2);c %= n2;//k1c *= n1 * d;//a = n1*k1+a1c += a1;nn = n1*n2*d;aa = c;return true;
}LL CRT(int len, LL *a, LL *n)
{LL a1 = a[0], n1 = n[0];LL a2, n2;for (int i = 1; i < len; i++){LL aa, nn;a2 = a[i], n2 = n[i];if (!Merge(a1, n1, a2, n2, aa, nn))return -1;a1 = aa;n1 = nn;}return (a1%n1 + n1) % n1;
}void work()
{LL ans, tmp = 1;memset(vis, false, sizeof(vis));int now = 0, step, t = 0;for (int i = 0; i < n; ++i){step = 0;for (int j = t%n; j < n; j = (j+1)%n){if (vis[j]) continue;step++;if (j == s[now]){now++;vis[j] = true;rest[i] = step%(n-i);mmod[i] = n-i;tmp = lcm(tmp, n-i);t = j;break;}}}ans = CRT(n, rest, mmod);if (ans == -1)printf("Creation August is a SB!\n");else{ans = (ans%tmp+tmp)%tmp;if (ans == 0) ans = tmp;cout << ans << endl;}
}int main()
{//freopen("test.in", "r", stdin);int T, u;scanf("%d", &T);for (int times = 1; times <= T; ++times){scanf("%d", &n);for (int i = 0; i < n; ++i){scanf("%d", &u);s[u-1] = i;}work();}return 0;
}