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 s;
long long a[200005];
long long compute(long long start, long long end, long long j) {
  long long n = end - start + 1;
  long long c1 = 0, c2 = 0, i;
  char ch = (char)('a' + j);
  if (start == end) {
    if (s[start] != ch)
      return 1;
    else
      return 0;
  }
  for (i = start; i < (start + n / 2); i++) {
    if (s[i] != ch) c1++;
  }
  for (i = (start + n / 2); i <= end; i++) {
    if (s[i] != ch) c2++;
  }
  long long ans1 = c1 + compute(start + n / 2, end, j + 1);
  long long ans2 = c2 + compute(start, start + n / 2 - 1, j + 1);
  return min(ans1, ans2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long test, ans, x, y, n, i, flag, j;
  cin >> test;
  while (test--) {
    cin >> n;
    x = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] == 1) x++;
    }
    long long turn = 0;
    if (x == n) {
      if (x % 2 == 0) {
        cout << "Second";
        goto l;
      } else {
        cout << "First";
        goto l;
      }
      continue;
    }
    for (i = 0; i < n; i++) {
      if (a[i] != 1) {
        if (turn % 2 == 0) {
          cout << "First";
          goto l;
        } else {
          cout << "Second";
          goto l;
        }
      } else {
        turn++;
      }
    }
  l:
    cout << "\n";
  }
}