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;
class DisjSet {
  int *rank, *parent, n;

 public:
  DisjSet(int n) {
    rank = new int[n];
    parent = new int[n];
    this->n = n;
    makeSet();
  }
  void makeSet() {
    for (int i = 0; i < n; i++) {
      parent[i] = i;
    }
  }
  int find(int x) {
    if (parent[x] != x) {
      parent[x] = find(parent[x]);
    }
    return parent[x];
  }
  void Union(int x, int y) {
    int xset = find(x);
    int yset = find(y);
    if (xset == yset) return;
    if (rank[xset] < rank[yset]) {
      parent[xset] = yset;
    } else if (rank[xset] > rank[yset]) {
      parent[yset] = xset;
    } else {
      parent[yset] = xset;
      rank[xset] = rank[xset] + 1;
    }
  }
};
int prefix_function(string s) {
  int n = (int)s.length();
  vector<int> pi(n);
  for (int i = 1; i < n; i++) {
    int j = pi[i - 1];
    while (j > 0 && s[i] != s[j]) j = pi[j - 1];
    if (s[i] == s[j]) j++;
    pi[i] = j;
  }
  return pi[n - 1];
}
long long power(long long x, long long y) {
  long long res = 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    int flag = 1, b = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] > 1) {
        if (b == 0)
          cout << "First"
               << "\n";
        else
          cout << "Second"
               << "\n";
        flag = 0;
        break;
      } else
        b = !b;
    }
    if (flag) {
      if (n % 2 == 0)
        cout << "Second"
             << "\n";
      else {
        cout << "First"
             << "\n";
      }
    }
  }
  return 0;
}