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 n, m, r, p, k, c, d;
long long *arr;
long long **mem;
bool play(long long i, long long player) {
  if (i == n - 1) {
    if (player == 1) {
      return true;
    } else
      return false;
  }
  if (mem[i][player + 1] != -1) {
    return mem[i][player + 1];
  }
  long long len = n - i;
  if (arr[i] == 1) {
    return play(i + 1, -player);
  }
  bool a = play(i + 1, player);
  bool b = play(i + 1, -player);
  bool ans;
  if (player == 1) {
    ans = a || b;
  } else {
    ans = a && b;
  }
  long long t = ans == true ? 1 : 0;
  mem[i][player + 1] = t;
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    cin >> n;
    arr = new long long[n];
    bool flag = true;
    for (long long i = 0; i < n; i++) {
      cin >> arr[i];
    }
    mem = new long long *[n];
    for (long long i = 0; i < n; i++) {
      mem[i] = new long long[3];
      for (long long j = 0; j < 3; j++) {
        mem[i][j] = -1;
      }
    }
    flag = play(0, 1);
    if (flag) {
      cout << "First\n";
    } else {
      cout << "Second\n";
    }
  }
}