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;
string to_string(basic_string<char>& x) { return "\"" + x + "\""; }
string to_string(char x) {
  string r = "";
  r += x;
  return "\'" + r + "\'";
}
string to_string(bool x) { return x ? "true" : "false"; }
template <typename T>
string to_str(T x) {
  return to_string(x);
}
template <typename T1, typename T2>
string to_str(pair<T1, T2> x) {
  return "(" + to_str(x.first) + ", " + to_str(x.second) + ")";
}
template <typename T>
string to_str(vector<T> x) {
  string r = "{";
  for (auto t : x) r += to_str(t) + ", ";
  return r.substr(0, r.length() - 2) + "}";
}
template <typename T1, typename T2>
string to_str(map<T1, T2> x) {
  string r = "{";
  for (auto t : x) r += to_str(t.first) + ": " + to_str(t.second) + ", ";
  return r.substr(0, r.length() - 2) + "}";
}
const unsigned long long MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int g, n, q;
  cin >> g;
  for (int t = 0; t < g; t++) {
    cin >> n;
    bool f = false;
    for (int i = 0; i < n; i++) {
      cin >> q;
      if (q > 1 && !f) {
        cout << (i % 2 == 0 ? "First\n" : "Second\n");
        f = true;
      }
    }
    if (!f) cout << (n & 1 ? "First\n" : "Second\n");
  }
}