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 en = "\n";
string sp = " ";
string t = "hello";
string Y = "YES\n";
string N = "NO\n";
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
bool isprm(LL a) {
  for (LL i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
bool palin(string a) {
  for (int i = 0; i < a.size(); i++) {
    if (a.at(i) != a.at(a.size() - i - 1)) return 0;
  }
  return 1;
}
bool isVowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
int toInt(string s) {
  int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
LL ceil2(LL a, LL b) {
  LL c = a / b;
  if (a % b) c++;
  return c;
}
LL bpow(LL a, LL b) {
  if (b == 0) return 1;
  LL r = bpow(a, b / 2);
  if (b % 2)
    return r * r * a;
  else
    return r * r;
}
LL binpow(LL a, LL b) {
  LL r = 1;
  while (b > 0) {
    if (b & 1) r = r * a;
    a = a * a;
    b >>= 1;
  }
  return r;
}
LL binpow(LL a, LL b, LL m) {
  a %= m;
  LL r = 1;
  while (b > 0) {
    if (b & 1) r = r * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return r;
}
template <typename T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
int main() {
  LL i, j, k, n, m, l, s = 0, x, y, tc = 1, p, q, a, b, c = 0;
  bool f = 0, ff = 0;
  string st, s1, s2, s3;
  char ch;
  cin >> tc;
  while (tc--) {
    cin >> n;
    vector<long long> v;
    for (i = 0; i < n; i++) {
      cin >> x;
      v.push_back(x);
      ;
    }
    c = f = 0;
    for (i = 0; i < n; i++) {
      if (v[i] == 1)
        c++;
      else {
        if (c % 2)
          cout << "Second" << en;
        else
          cout << "First" << en;
        f = 1;
        break;
      }
    }
    if (f) continue;
    if (c % 2)
      cout << "First" << en;
    else
      cout << "Second" << en;
  }
}