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>
using namespace std;
vector<long long int> primes;
void SOE(long long int n) {
  primes.erase(primes.begin(), primes.end());
  bool prime[n + 1];
  memset(prime, true, sizeof(prime));
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  for (long long int p = 2; p <= n; p++)
    if (prime[p]) {
      primes.push_back(p);
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int i, c = 0, j, k, t, f = 0, q, n, f1 = 0, y, x, m, l = 0, r, p;
  string s, s1, s2;
  cin >> t;
  while (t--) {
    cin >> n;
    long long int a[n];
    for (i = 0; i < n; i++) cin >> a[i];
    x = 0;
    f = 0;
    c = 0;
    for (i = 0; i < n; i++) {
      if (a[i] == 1)
        c++;
      else {
        break;
      }
    }
    if (count(a, a + n, 1) == n) {
      if (n % 2 == 0)
        cout << "Second\n";
      else
        cout << "First\n";
    } else {
      if (c % 2 != 1)
        cout << "First\n";
      else
        cout << "Second\n";
    }
  }
}