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;
long long int bigmod(long long int a, long long int b) {
  if (b == 0) {
    return 1;
  }
  long long int tm = bigmod(a, b / 2);
  tm = (tm * tm) % 1000000007;
  if (b % 2 == 1) tm = (tm * a) % 1000000007;
  return tm;
}
int egcd(int a, int b, int &x, int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  int x1, y1;
  int d = egcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cout << setprecision(10) << fixed;
  ;
  solve();
  return 0;
}
void solve() {
  int t;
  cin >> t;
  for (int i = 0; i < int(t); i++) {
    int n;
    cin >> n;
    int a[n];
    int ind = n, f = 0, c = 0, f1 = 1;
    for (int j = 0; j < int(n); j++) {
      cin >> a[j];
      if (a[j] == 1 && f == 0)
        c++;
      else
        f = 1;
    }
    if (c == 0 || (c % 2 == 0 && c != n) || (c % 2 == 1 && c == n))
      cout << "First" << endl;
    else
      cout << "Second" << endl;
  }
}