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;
long long setBit(long long mask, int p) { return (mask | (1 << p)); }
long long resetBit(long long mask, int p) { return (mask & ~(1 << p)); }
long long flipBit(long long mask, int p) { return (mask ^ (1 << p)); }
bool check(long long mask, int p) { return (mask & (1 << p)); }
int msb(long long N) { return N ? 63 - __builtin_clzll(N) : -1; }
int lsb(long long N) { return __builtin_ffs(N) - 1; }
int countOn(long long mask) { return __builtin_popcountll(mask); }
int countOff(long long mask) { return msb(mask) - countOn(mask) + 1; }
long long bigmod(long long a, long long b, long long mod) {
  a %= mod;
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return ret;
}
long long modInverse(long long a) { return bigmod(a, 1, 1000000007 - 2); }
mt19937_64 mt_rnd_64(chrono::steady_clock::now().time_since_epoch().count());
long long rnd(long long l, long long r) {
  return (mt_rnd_64() % (r - l + 1)) + l;
}
long long arr[100009];
int main() {
  int t, n, i;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    int ones = 0;
    for (i = 0; i < n; i++) {
      scanf("%d", &arr[i]);
      ones += (arr[i] == 1);
    }
    if (ones == n) {
      if (n % 2)
        printf("First\n");
      else
        printf("Second\n");
    } else {
      int pos = 0;
      while (arr[pos] == 1) {
        pos++;
      }
      if (pos % 2 == 0)
        printf("First\n");
      else
        printf("Second\n");
    }
  }
}