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;
template <typename T>
inline void INT(T &n) {
  n = 0;
  int f = 1;
  int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
template <typename T>
T gcd(T a, T b) {
  return !b ? a : gcd(b, a % b);
}
template <typename T>
inline void umin(T &a, T b) {
  a = a < b ? a : b;
}
template <typename T>
inline void umax(T &a, T b) {
  a = a > b ? a : b;
}
template <typename T, typename W>
inline void INT(T &x, W &y) {
  INT(x), INT(y);
}
template <typename T, typename W, typename Q>
inline void INT(T &x, W &y, Q &z) {
  INT(x, y), INT(z);
}
const int N = 1e5 + 7;
const int inf = 1e9 + 7;
int solve() {
  int n;
  INT(n);
  vector<int> a(n);
  for (int &x : a) {
    INT(x);
  }
  string finisher = "second";
  for (int i = n - 2; i > 0; --i) {
    if (finisher == "second") {
      if (a[i] == 1) {
        finisher = "first";
      }
    } else {
      finisher = "second";
    }
  }
  if (n == 1 | finisher == "first" || (finisher == "second" && a[0] > 1)) {
    printf("First\n");
  } else {
    printf("Second\n");
  }
  return 0;
}
int main() {
  int tests = 1, CaseNo = 0;
  INT(tests);
  while (tests--) {
    solve();
  }
  return 0;
}