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;
template <class T, class U>
void umin(T& x, const U& y) {
  x = min(x, (T)y);
}
template <class T, class U>
void umax(T& x, const U& y) {
  x = max(x, (T)y);
}
template <class T, class U>
void init(vector<T>& v, U x, size_t n) {
  v = vector<T>(n, x);
}
template <class T, class U>
void init(vector<T>& v, U x, size_t n, size_t m...) {
  v = vector<T>(n);
  for (auto a : v) init(a, x, m);
}
int main(int argc, char** argv) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(15);
  if (argc == 2 && atoi(argv[1]) == 123456789)
    freopen("d:\\code\\cpp\\contests\\stdin", "r", stdin);
  vector<string> ss = {"First", "Second"};
  int T;
  cin >> T;
  for (int(tt) = 0; (tt) < (T); (tt)++) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int(i) = 0; (i) < (n); (i)++) cin >> a[i];
    int w = (n + 1) % 2;
    for (int(i) = 0; (i) < (n); (i)++) {
      if (a[i] > 1) {
        w = i % 2;
        break;
      }
    }
    cout << ss[w] << endl;
  }
  if (argc == 2 && atoi(argv[1]) == 123456789)
    cout << clock() * 1.0 / CLOCKS_PER_SEC << " sec\n";
  return 0;
}