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. 
#include <bits/stdc++.h>
using namespace std;
bool isPrime[1000005] = {false};
vector<long long> prm;
vector<long long> SPF(1000005);
void sieve() {
  isPrime[1] = isPrime[0] = 0;
  long long i;
  for (i = 2; i < 1000005; ++i) isPrime[i] = 1;
  for (i = 2; i * i < 1000005; ++i) {
    if (isPrime[i]) {
      for (long long j = i * i; j < 1000005; j += i) isPrime[j] = 0;
    }
  }
  if (!prm.empty()) prm.clear();
  for (i = 2; i < 1000005; ++i) {
    if (isPrime[i]) prm.push_back(i);
  }
}
bool flip(bool f) {
  if (f) return 0;
  return 1;
}
long long fastPower(long long a, long long b) {
  if (b == 0) return 1;
  long long prod = 1;
  while (b > 0) {
    if (b & 1) prod = prod * a;
    b = b >> 1;
    a = a * a;
  }
  return prod;
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; i++) {
    res *= i;
  }
  return res;
}
long long nCr(long long n, long long r) {
  return (long double)fact(n) / (long double)(fact(r) * fact(n - r));
}
void solve() {
  long long n, i;
  cin >> n;
  vector<long long> a(n);
  for (i = 0; i < n; ++i) cin >> a[i];
  for (i = 0; i < n; ++i) {
    if (a[i] != 1) break;
  }
  string ans = ((i == n) ^ (i & 1)) ? "Second" : "First";
  cout << ans << "\n";
}
int main() {
  int tc = 1;
  scanf("%d", &tc);
  while (tc--) {
    solve();
  }
  return 0;
}
