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("O2")
using namespace std;
const long long int N = 1e9 + 7;
long long int pow(long long int x, long long int y) {
  long long int res = 1;
  while (y > (long long int)0) {
    if (y & (long long int)1) res = ((res) * (x));
    y = y >> (long long int)1;
    x = ((x) * (x));
  }
  return res;
}
long long int powm(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > (long long int)0) {
    if (y & (long long int)1) res = (res * x) % p;
    y >>= (long long int)1;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long int i, j, k, ans, a, b, c, mx, mn, t, n, m, x, temp, p, q, y, z, l,
      r, g, h, flg, sum, d, odd, even, prev, cnt, diff;
  cin >> n;
  vector<long long int> arr(n);
  for (i = 0; i < n; i++) cin >> arr[i];
  x = 0;
  for (i = 0; i < n; i++) {
    x ^= 1;
    if (arr[i] != 1) {
      if (x == 1)
        cout << "First" << endl;
      else
        cout << "Second" << endl;
      return;
    }
  }
  if (x == 1)
    cout << "First" << endl;
  else
    cout << "Second" << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
}