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 power(long long a, long long n, long long m) {
  long long res = 1;
  while (n != 0) {
    if (n % 2 == 1) {
      res = (res * a) % m;
      n--;
    } else {
      a = (a * a) % m;
      n = n / 2;
    }
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    long long ones = 0;
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
      if (arr[i] == 1) {
        ones++;
      }
    }
    long long one = 0, two = 0, c = 0;
    long long prefix[n];
    prefix[n - 2] = 0;
    for (long long i = n - 2; i >= 0; i--) {
      if (arr[i] == 1) {
        c++;
        if (i - 1 >= 0) {
          prefix[i - 1] = c;
        }
      } else {
        if (i - 1 >= 0) {
          prefix[i - 1] = 0;
        }
        c = 0;
      }
    }
    long long flip = 0;
    for (long long i = 0; i < n - 1; i++) {
      if (arr[i] == 1) {
        flip = flip ^ 1;
      }
      if (arr[i] != 1) {
        if (prefix[i] % 2 == 1) {
          flip = flip ^ 1;
        } else {
          flip = flip ^ 0;
        }
      } else {
      }
    }
    if (flip == 0) {
      cout << "First"
           << "\n";
    } else {
      cout << "Second"
           << "\n";
    }
  }
  return 0;
}