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 int lastR = 1e5 + 1;
int n, nTask, arr[lastR << 1], dp2[lastR << 1][3];
bool dp[lastR << 1][3];
inline int retMin(int x, int y) {
  if (x > y)
    return y;
  else
    return x;
}
inline int retMax(int x, int y) {
  if (x > y)
    return x;
  else
    return y;
}
bool rec(int i, bool flag) {
  if (i > n) return flag;
  bool &ret = dp[i][flag];
  if (dp2[i][flag] == nTask) return ret;
  dp2[i][flag] = nTask;
  if (!flag) {
    if (arr[i] == 1) {
      ret = rec(i + 1, !flag);
    } else {
      ret = retMax(rec(i + 1, !flag), rec(i + 1, flag));
    }
  } else {
    if (arr[i] == 1) {
      ret = rec(i + 1, !flag);
    } else {
      ret = retMin(rec(i + 1, !flag), rec(i + 1, flag));
    }
  }
  return ret;
}
int main() {
  int cases;
  scanf("%d", &cases);
  while (cases--) {
    nTask++;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
    bool ans = rec(1, 0);
    if (ans)
      printf("First\n");
    else
      printf("Second\n");
  }
  return 0;
}