#include <bits/stdc++.h>
// #if __has_include(<atcoder/all>)
// #include <atcoder/all>
// using namespace atcoder;
// #endif
using namespace std;

// #include "boost/multiprecision/cpp_int.hpp"
// using namespace boost::multiprecision;

// #pragma GCC target("avx2")
// #pragma GCC optimize("O3")
// #pragma GCC optimize("unroll-loops")

#define ll long long
#define rep(i, n) for (ll i = 0; i < n; ++i)
#define rep_up(i, a, n) for (ll i = a; i < n; ++i)
#define rep_down(i, a, n) for (ll i = a; i >= n; --i)
#define P pair<ll, ll>
#define pb push_back
#define bit_count(x) __builtin_popcountll(x)
#define gcd(a, b) __gcd(a, b)
#define lcm(a, b) a / gcd(a, b) * b
// #define endl "\n"

#define all(v) v.begin(), v.end()
#define fi first
#define se second
#define vll vector<ll>
#define vvll vector<vll>
#define vvvll vector<vvll>
#define vvvvll vector<vvvll>
#define pqll priority_queue<ll>
#define pqllg priority_queue<ll, vector<ll>, greater<ll>>

template <class T>
inline void vin(vector<T> &v) { rep(i, v.size()) cin >> v.at(i); }
template <class T>
using V = vector<T>;

constexpr ll INF = (1ll << 60);
// constexpr ll mod = 1000000007;
// constexpr ll mod = 998244353;

// using mint = atcoder::modint;
// using mint = atcoder::modint998244353;
// using mint = atcoder::modint1000000007;
#define vmint vector<mint>
#define vvmint vector<vmint>
#define vvvmint vector<vvmint>
#define vvvvmint vector<vvvmint>

constexpr double pi = 3.14159265358979323846;

random_device seed_gen;
mt19937 engine(seed_gen());

template <typename T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <typename T>
void pt(T val)
{
    cout << val << "\n";
}
template <typename T>
void pt_vll(vector<T> &v)
{
    ll vs = v.size();
    rep(i, vs)
    {
        cout << v[i];

        if (i == vs - 1)
            cout << "\n";
        else
            cout << " ";
    }
}
template <typename T>
void pt_vmint(vector<T> &v)
{
    ll vs = v.size();
    rep(i, vs)
    {
        cout << v[i].val();

        if (i == vs - 1)
            cout << "\n";
        else
            cout << " ";
    }
}
ll mypow(ll a, ll n)
{
    ll ret = 1;
    if (n == 0)
        return 1;
    if (a == 0)
        return 0;
    rep(i, n)
    {
        if (ret > (ll)(9e18 + 10) / a)
            return -1;
        ret *= a;
    }
    return ret;
}
long long modpow(long long a, long long n, long long mod)
{
    long long res = 1;
    while (n > 0)
    {
        if (n & 1)
            res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
long long modinv(long long a, long long m)
{
    long long b = m, u = 1, v = 0;
    while (b)
    {
        long long t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}

void solve();
// ここまでテンプレ

ll sum_N = 0;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    // cout << fixed << setprecision(16);
    ll T;
    cin >> T;
    rep(ca, T)
        solve();
    ll x;
    while (cin >> x)
    {
        assert(false);
    }
    assert(sum_N <= 200000);
}

bool sub_solve(ll n, vll &p)
{
    rep_up(i, 1, n)
    {
        assert(1 <= p[i] && p[i] - 1 < i + 1);
    }
    vll a(n);
    rep(i, n - 1)
    {
        ll x = p[i + 1] - 1;
        a[x]++;
    }
    bool flag = false; // 先手勝ちかどうか
    rep_up(i, 1, n)
    {
        if (a[i] == 0)
        {
            ll x = i;
            ll cnt = 0;
            while (a[x] <= 1 && x != 0)
            {
                cnt++;
                x = p[x] - 1;
                // cerr << x << endl;
            }
            // cerr << i + 1 << " " << cnt << endl;
            if (cnt % 2 == 1)
            {
                flag = true;
                // break;
            }
        }
    }
    return flag;
}

void solve()
{
    ll n, m, k, cnt = 0, sum = 0, ans = 0;
    cin >> n;
    assert(2 <= n && n <= 200000);
    sum_N += n;
    vll p(n); // 親となる頂点番号, 根は -1
    rep(i, n - 1)
    {
        cin >> p[i + 1];
    }
    bool flag = sub_solve(n, p);
    if (flag)
    {
        pt("First");
    }
    else
    {
        pt("Second");
    }
}