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;
void wenjian() {
  freopen("concatenation.in", "r", stdin);
  freopen("concatenation.out", "w", stdout);
}
void tempwj() {
  freopen("hash.in", "r", stdin);
  freopen("hash.out", "w", stdout);
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long qpow(long long a, long long b, long long mod) {
  a %= mod;
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ans;
}
struct cmp {
  bool operator()(const pair<int, long long>& a,
                  const pair<int, long long>& b) {
    return a.second > b.second;
  }
};
int lb(int x) { return x & -x; }
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int M = 4e4 + 2;
const long long mod = 80112002;
const int maxn = 2e5 + 5 + 7;
int a[maxn];
int b[maxn];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    int f = 1;
    for (int i = n - 1; i >= 1; i--) {
      if (a[i] != 1) {
        f = 1;
      } else
        f ^= 1;
    }
    if (f)
      printf("First\n");
    else
      printf("Second\n");
  }
}