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;
const long long INF = 2000000000000000000LL;
const int inf = 0x3f3f3f3f;
long long MOD = 1e9 + 7;
const long double EPS = 1e-9;
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline double my_sqrt(T n) {
  double high = n, low = 0, mid, ans;
  int cnt = 100;
  while (cnt--) {
    mid = low + (high - low) / 2;
    if (mid * mid <= n)
      ans = mid, low = mid;
    else
      high = mid;
  }
  return ans;
}
template <class T>
inline T bigmod(T b, T p) {
  if (p <= 0) return 1;
  if (p & 1) return (b * bigmod(b, p - 1)) % MOD;
  long long x = bigmod(b, p >> 1);
  return (x * x) % MOD;
}
const int maxn = 1e5;
int arr[maxn + 5], n;
int main() {
  int test;
  scanf("%d", &test);
  ;
  while (test--) {
    scanf("%d", &n);
    ;
    int indx = -1, ck = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &arr[i]);
      ;
      if (arr[i] == 1 && ck == 0) {
        indx = i;
      } else
        ck = 1;
    }
    if (indx == -1 || indx == n - 1) {
      if (indx == -1)
        puts("First");
      else if (n % 2)
        puts("First");
      else
        puts("Second");
    } else {
      if (indx % 2)
        puts("First");
      else
        puts("Second");
    }
  }
  return 0;
}