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 binarySearch(vector<int>* arr, int* num, int* low, int* mid, int* high) {
  while (true) {
    if (arr->at(*mid) <= *num && arr->at(*mid + 1) > *num) {
      break;
    } else if (*num > arr->at(*mid)) {
      *low = *mid;
      *mid = (*low + *high) / 2;
    } else if (*num < arr->at(*mid)) {
      *high = *mid;
      *mid = (*high + *low) / 2;
    }
  }
}
void gcd(int* a, int* b, int* _gcd) {
  if (*a == *b) {
    *_gcd = *a;
  } else if ((*a % 2 == 0 && *b % 2 != 0) || (*a % 2 != 0 && *b % 2 == 0)) {
    *_gcd = 1;
  } else if (*a == 0 || *b == 0) {
    *_gcd = 1;
  } else if (*a > *b) {
    for (int i = 2; i < *a / 2; i++) {
      if (*a % i == 0 && *b % i == 0) {
        *_gcd *= i;
        *a /= i;
        *b /= i;
        i--;
      }
      if (*a == 1 || *b == 1) {
        break;
      }
    }
  } else if (*b > *a) {
    for (int i = 2; i < *b / 2; i++) {
      if (*a % i == 0 && *b % i == 0) {
        *_gcd *= i;
        *a /= i;
        *b /= i;
        i--;
      }
      if (*a == 1 || *b == 1) {
        break;
      }
    }
  }
}
int main() {
  int* t = new int;
  cin >> *t;
  for (int h = 0; h < *t; h++) {
    int* n = new int;
    cin >> *n;
    int* firstSecond = new int(-1);
    bool* g = new bool(false);
    long long* temp = new long long;
    string* winner = new string;
    for (int i = 0; i < *n; i++) {
      cin >> *temp;
      if (*temp == 1) {
        *firstSecond *= -1;
      }
      if (*temp > 1 && winner->empty()) {
        if (*firstSecond == -1) {
          *winner = "First";
        } else {
          *winner = "Second";
        }
      }
    }
    if (winner->empty()) {
      if (*firstSecond == -1) {
        *winner = "Second";
      } else {
        *winner = "First";
      }
    }
    cout << *winner << endl;
    delete temp, n, firstSecond, g, winner;
  }
  delete t;
}