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. 
import sys
import math
input = sys.stdin.readline
from functools import cmp_to_key;

def pi():
    return(int(input()))
def pl():
    return(int(input(), 16))
def ti():
    return(list(map(int,input().split())))
def ts():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
f = [];
def factMod(n,m):
    global f;
    f = [1 for i in range(n+1)];
    f[0] = 1;
    for i in range(1,n+1):
        f[i] = (f[i-1]*i)%m;

def fact(n):
    global f;
    f = [1 for i in range(n)];
    for i in range(1,n):
        f[i] = f[i-1]*i;

def fast_mod_exp(a,b,m):
    res = 1;
    while b > 0:
        if b & 1:
            res = (res*a)%m;
        a = (a*a)%m;
        b = b >> 1;
    return res;

def inverseMod(n,m):
    return fast_mod_exp(n,m-2,m);

def ncrMod(n,r,m):
    if r == 0: return 1;
    if n < r: return 0;
    return ((f[n]*inverseMod(f[n-r],m))%m*inverseMod(f[r],m))%m;

def ncr(n,r):
    if r == 0: return 1;
    if n < r: return 0;
    return (f[n])/(f[r]*f[n-r]);
def nc2(n):
    return (n*(n-1))/2;

def main():
    B();

def A():
    t = pi();
    while t:
        t -= 1;
        [n,m] = ti();
        a = ti();
        b = ti();
        mpa = [0 for i in range(1005)];
        mpb = [0 for i in range(1005)];
        for i in range(n):
            mpa[a[i]] = 1;
        for i in range(m):
            mpb[b[i]] = 1;
        f = 0;
        for i in range(1005):
            if mpa[i] == 1 and mpb[i] == 1:
                print('YES');
                print(1,i);
                f = 1;
                break;

        if f == 0:
            print('NO');

def B():
    t = pi();
    while t:
        t -= 1;
        n = pi();
        a = ti();
        if n == 1:
            print('First');
            continue;
        isAllOne = 1;
        for i in range(n):
            if a[i] != 1:
                isAllOne = 0;
                break;

        if isAllOne:
            if n%2 == 0: print('Second');
            else: print('First');
            continue;
        if a[0] > 1:
            print('First');
            continue;
        turn = 1;
        for i in range(n):
            if a[i] > 1:
                if turn == 1: print('First');
                else: print('Second');
                break;
            else:
                turn = 1 if turn == 2 else 2;
         


main();