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. 
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
#pragma optimize("-O3")
inline long long read() {
  long long x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x = f ? -x : x;
}
inline void write(long long x, bool f) {
  if (x == 0) {
    putchar('0');
    if (f) putchar('\n');
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  static char s[23];
  int l = 0;
  while (x != 0) s[l++] = x % 10 + 48, x /= 10;
  while (l) putchar(s[--l]);
  if (f) putchar('\n');
}
int lowbit(int x) { return x & (-x); }
template <class T>
T big(const T &a1, const T &a2) {
  return a1 > a2 ? a1 : a2;
}
template <class T>
T sml(const T &a1, const T &a2) {
  return a1 < a2 ? a1 : a2;
}
template <typename T, typename... R>
T big(const T &f, const R &...r) {
  return big(f, big(r...));
}
template <typename T, typename... R>
T sml(const T &f, const R &...r) {
  return sml(f, sml(r...));
}
void debug_out() { cout << '\n'; }
template <typename T, typename... R>
void debug_out(const T &f, const R &...r) {
  cout << f << " ";
  debug_out(r...);
}
const char ptout[] = "%d";
template <typename T>
void print(const T &f) {
  printf(ptout, f);
  putchar('\n');
}
template <typename T, typename... R>
void print(const T &f, const R &...r) {
  printf(ptout, f);
  putchar(' ');
  print(r...);
}
const int HMOD[] = {1000000009, 1004535809};
const long long BASE[] = {1572872831, 1971536491};
const long long INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
const int MOD = 1e9 + 7;
const int MXN = 2e5 + 5;
const int MXE = 2e6 + 6;
int n, m, k;
int ar[MXN], dp[MXN];
int main() {
  int tim;
  scanf("%d", &tim);
  while (tim--) {
    n = read();
    int x, ans = 0, w = 0, g = -1;
    for (int i = 1; i <= n; ++i) {
      x = read();
      ar[i] = x;
      if (w == 0 && x == 1) {
        g = i;
      }
      if (x != 1) w = 1;
    }
    if (g == -1)
      ans = 1;
    else if (g == n)
      ans = g % 2;
    else
      ans = g % 2 == 0;
    if (ans)
      printf("First\n");
    else
      printf("Second\n");
  }
  return 0;
}
