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;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int f = 1;
    int s = 0;
    vector<int> A;
    for (int i = 0; i < n; i++) {
      int a;
      cin >> a;
      A.push_back(a);
    }
    if (A.size() == 1) {
      cout << "First" << endl;
      continue;
    }
    if (n == 2) {
      if (A[0] != 1) {
        cout << "First" << endl;
        continue;
      } else {
        cout << "Second" << endl;
        continue;
      }
    }
    for (int i = 0; i < n - 2; i++) {
      if (A[i] != 1 && A[i + 1] != 1) {
        if (f == 1) {
          f = 1;
          s = 0;
        } else {
          s = 1;
          f = 0;
        }
      } else if (A[i] != 1 && A[i + 1] == 1) {
        int count = 0;
        int j = i + 1;
        while (A[j] == 1 && j <= n - 2) {
          count++;
          j++;
        }
        if (count % 2 == 1) {
          if (f == 1) {
            s = 1;
            f = 0;
          } else {
            f = 1;
            s = 0;
          }
        } else {
          if (f == 1) {
            f = 1;
            s = 0;
          } else {
            s = 1;
            f = 0;
          }
        }
      } else if (A[i] == 1) {
        if (f == 1) {
          s = 1;
          f = 0;
        } else {
          f = 1;
          s = 0;
        }
      }
    }
    if (A[n - 2] != 1 && f == 1) {
      cout << "First" << endl;
      continue;
    } else if (A[n - 2] == 1 && f == 1) {
      cout << "Second" << endl;
      continue;
    } else if (A[n - 2] != 1 && s == 1) {
      cout << "Second" << endl;
      continue;
    } else if (A[n - 2] == 1 && s == 1) {
      cout << "First" << endl;
      continue;
    }
  }
  return 0;
}