Problem: There are n piles of stones, where the i-th pile has a_i stones. Two people play a game, where they take alternating turns removing stones.

In a move, a player may remove a positive number of stones from the first non-empty pile (the pile with the minimal index, that has at least one stone). The first player who cannot make a move (because all piles are empty) loses the game. If both players play optimally, determine the winner of the game.

Input

The first line contains a single integer t (1≤ t≤ 1000) — the number of test cases. Next 2t lines contain descriptions of test cases.

The first line of each test case contains a single integer n (1≤ n≤ 10^5) — the number of piles.

The second line of each test case contains n integers a_1,…,a_n (1≤ a_i≤ 10^9) — a_i is equal to the number of stones in the i-th pile.

It is guaranteed that the sum of n for all test cases does not exceed 10^5.

Output

For each test case, if the player who makes the first move will win, output "First". Otherwise, output "Second".

Example

Input


7
3
2 5 4
8
1 1 1 1 1 1 1 1
6
1 2 3 4 5 6
6
1 1 2 1 2 2
1
1000000000
5
1 2 2 1 1
3
1 1 1


Output


First
Second
Second
First
First
Second
First

Note

In the first test case, the first player will win the game. His winning strategy is: 

  1. The first player should take the stones from the first pile. He will take 1 stone. The numbers of stones in piles will be [1, 5, 4]. 
  2. The second player should take the stones from the first pile. He will take 1 stone because he can't take any other number of stones. The numbers of stones in piles will be [0, 5, 4]. 
  3. The first player should take the stones from the second pile because the first pile is empty. He will take 4 stones. The numbers of stones in piles will be [0, 1, 4]. 
  4. The second player should take the stones from the second pile because the first pile is empty. He will take 1 stone because he can't take any other number of stones. The numbers of stones in piles will be [0, 0, 4]. 
  5. The first player should take the stones from the third pile because the first and second piles are empty. He will take 4 stones. The numbers of stones in piles will be [0, 0, 0]. 
  6. The second player will lose the game because all piles will be empty. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")
#pragma GCC optimize("Os")
using namespace std;
using ll = long long int;
using ld = long double;
using pll = pair<ll, ll>;
using pii = pair<int, int>;
using vll = vector<ll>;
using vi = vector<int>;
const ll mod = (ll)(1e9) + 7LL;
const ll M = 988244353LL;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};
const ld pi = acos(-1);
ll gcd(ll a, ll b) { return (b ? gcd(b, a % b) : a); }
ll P(ll B, ll power, ll modulo) {
  ll ans = 1LL;
  while (power > 0LL) {
    if (power % 2LL == 1LL) {
      ans = (ans * B) % modulo;
    }
    B = (B * B) % modulo;
    power /= 2LL;
  }
  return ans;
}
bool isPrime(ll n) {
  if (n <= 1LL) {
    return false;
  }
  if (n <= 3LL) {
    return true;
  }
  if (n % 2 == 0LL || n % 3 == 0LL) {
    return false;
  }
  for (ll i = 5LL; (i * i) <= n; i += 6LL) {
    if (n % i == 0LL || n % (i + 2LL) == 0LL) {
      return false;
    }
  }
  return true;
}
void vok() {
  ios_base::sync_with_stdio(false);
  cout.tie(NULL);
  cin.tie(NULL);
}
const int mxN = int(1e5) + 100;
void solve() {
  int n;
  cin >> n;
  vll a(n);
  for (auto &x : a) {
    cin >> x;
  }
  int pick = 0;
  vi pre(n, 0);
  for (ll i = 0; i < n; i++) {
    if (i) {
      pre[i] = pre[i - 1] + (a[i] != 1);
    } else {
      pre[i] = (a[i] != 1);
    }
  }
  for (ll i = 0; i < n - 1; i++) {
    if (a[i] == 1) {
      pick ^= 1;
    } else {
      break;
    }
  }
  if (pick == 0) {
    cout << "First"
         << "\n";
  } else {
    cout << "Second"
         << "\n";
  }
}
int main() {
  vok();
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}